# Factorization versus primality testing

Let $n$ be a large positive integer whose “prime versus composite” status is not known. One way to know whether $n$ is prime or composite is to factor $n$ into its prime factors. If there is a non-trivial factor (one that is neither 1 nor $n$), it is composite. Otherwise $n$ is prime. This may sound like a reasonable approach in performing primality testing – checking whether a number is prime or composite. In reality, factoring and primality testing, though related, are very different problems. For a very large number (e.g. with at least 300 decimal digits), it is possible that, even with the state of the art in computing, factoring it may take more than a few million years. On the other hand, it will take a modern computer less than a second to determine whether a 300-digit number is prime or composite. Interestingly this disparity is one reason that makes the RSA work as a practical and secure cryptosystem. In this post, we use the RSA cryptosystem as an example to give a sense that factoring is a “hard” problem while primality testing is an “easy” problem. The primality test used in the examples is the Fermat primality test.

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The brute force approach

There is a natural and simple approach in factoring, which is to do trial divisions. To factor the number $n$, we divide $n$ by every integer $a$ in the range $1. Once a factor $a$ is found, we repeat the process with the complementary factor $\frac{n}{a}$ until all the prime factors of $n$ are found. This is simple in concept and is sure to produce the correct answer. For applications in cryptography, this brute force approach is essentially useless since the amount of time to try every candidate factor is prohibitively huge. The amount of time required may be more than the age of the universe if the brute force approach is used.

The brute force approach can be improved upon slightly by doing the trial divisions using candidate factors up to $\sqrt{n}$. It is well known that if a composite integer $n$ is greater than one, then it has a prime divisor $d$ such that $1. So instead of dividing $n$ by every number $a$ with $1, we can divide $n$ by every prime number $a$ with $1. But even this improved brute force approach will still take too long to be practical.

Let’s look at a quick example for brute force factoring. Let $n=96638243$. Note that $\sqrt{n}=\sqrt{96638243}=9676$. There are 1192 odd primes less than 9676. In dividing $n$ by these primes, we stop at 127 and we have $n=96638243=127 \cdot 737309$. We now focus the attention on $737309$. Note that $\sqrt{737309}=858.67$ and there are 147 odd primes less than 858. Dividing $737309$ by these 147 candidate factors, we find that none of them is a factor. We can conclude $737309$ is prime. Then we have the factorization $n=96638243=127 \cdot 737309$.

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Example of RSA

RSA is a public-key cryptosystem and is widely used for secure data transmission. The RSA public key consists of two parts. One is the modulus that is the product of two distinct prime factors. Suppose the modulus is called $N$ and we have $N=pq$ where $p$ and $q$ are distinct prime numbers. How large does $N$ have to be? The larger the $N$ is, the more secure RSA is. The current practice is that for corporate use the modulus is at least a 1024-bit number (the bit size is called the key length). If data is extra sensitive or if the data needs to be retained for a long time, then a larger key length should be used (e.g. 2048-bit). With a 1024-bit modulus $N=pq$, each prime factor is a 512-bit number. The other part of the RSA public key is the encryption key, which is an integer $e$ that is relatively prime to the integer $(p-1) \cdot (q-1)$.

Let’s say we want to generate a 1024-bit modulus. There are two challenges with a key of this size. One is that a reliable way is needed to obtain two prime numbers that are 512-bit long. Given a large integer that is at least 512-bit long, how do we determine reliably that it is prime? Is it possible to factor a 512-bit integer using the brute force approach? The other challenge is from the perspective of an attacker – successful factoring the 1024-bit modulus would break RSA and allow the attacker to read the secret message. Let’s look at the problem of the attacker trying to factor a 1024-bit number. A 1024-bit number is approximately $2^{1024}$. The following calculation converts it to a decimal basis:

$\displaystyle 2^{1024}=(10^{\text{log(2)}})^{1024} \approx 10^{308.25}$

We use $\text{log}(x)$ to denote the logarithm of base 10. Note that $1024 \cdot \text{log}(2)=308.25$. So a 1024-bit number has over 300 digits.

Let’s see what the challenge is if you want to factor a 1024-bit number. Suppose your chosen large number $n$ is such that $n \approx 10^{308}$. Note that $\sqrt{10^{308}}=10^{154}$. According to the improved brute force approach described above, in effect you will need to divide $n$ by every prime number less than $10^{154}$.

Now let’s get an estimate on the number of prime numbers less than $10^{154}$. According to the prime number theorem, the number of prime numbers at most $x$ is approximately

$\displaystyle \pi(x) \approx \frac{x}{\text{ln}x}$

where $\pi(x)$ is the number of primes at most $x$. Then $\pi(10^{154}) \approx 2.82 \cdot 10^{151}$. This is certainly a lot of prime numbers to check.

It is hard to comprehend such large numbers. Let’s put this into perspective. Currently the world population is about 7 billion. Let’s say each person in the world possesses a supercomputer that can check $10^{40}$ prime numbers per second (i.e. to check whether they are factors of the number $n$). This scenario clearly far exceeds the computing resources that are currently available. Suppose that the 7 billion supercomputers are available and that each one can check $10^{40}$ many primes per second. Then in each second, the following is the number of prime numbers that can be checked by the 7 billion supercomputers.

$\displaystyle 7 \cdot 10^9 \cdot 10^{40}=7 \cdot 10^{49} \text{ prime numbers per second}$

The following is the number of seconds it will take to check $2.82 \cdot 10^{151}$ many prime numbers:

$\displaystyle \frac{2.82 \cdot 10^{151}}{7 \cdot 10^{49}} \approx 4 \cdot 10^{101} \text{ seconds}$

The universe is estimated to be about 13 billion years old. The following calculation converts it to seconds.

$13 \text{ billion years}=13 \cdot 10^9 \cdot 365 \cdot 24 \cdot 3600 \approx 4 \cdot 10^{17} \text{ seconds}$

With 7 billion fast suppercomputers (one for each person in the world) running in the entire life of the universe, you can only finish checking

$\displaystyle \frac{4 \cdot 10^{17}}{4 \cdot 10^{101}}=\frac{1}{10^{84}}$

of the $2.82 \cdot 10^{151}$ many prime numbers. Note that $\frac{1}{10^{84}}$ is a tiny portion of 1%. So by taking the entire life of the universe to run the 7 billion supercomputers, each checking $10^{40}$ many candidate prime factors per second, you would not even make a dent in the problem!

The security of RSA rests on the apparent difficulty of factoring large numbers. If the modulus $N=pq$ can be factored, then an eavesdropper can obtain the private key from the public key and be able to read the message. The difficulty in factoring means there is a good chance that RSA is secure. In order to break RSA, an attacker would probably have to explore other possible vulnerabilities instead of factoring the modulus.

By carrying out a similar calculation, we can also see that factoring a 512-bit number by brute force factoring is also not feasible. Thus in the RSA key generation process, it is not feasible to use factoring as a way to test primality. The alternative is to use efficient primality tests such as Fermat test or Miller-Rabin test. The computation for these tests is based on the fast powering algorithm, which is a very efficient algorithm.

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The story told by RSA numbers

The required time of more than the life of the universe as discussed above is based on the naïve brute force approach of factoring. There are many other factoring approaches that are much more efficient and much faster, e.g., the quadratic sieve algorithm, the number field sieve algorithm, and the general number field sieve algorithm. For these methods, with ample computing resources at the ready, factoring a 1024-bit or 2048-bit number may not take the entire life of the universe but make take decades or more. Even with these better methods, the disparity between slow factoring and fast primality testing is still very pronounced and dramatic.

The best evidence of slow factoring even with using modern methods is from the RSA numbers. The RSA numbers are part of the the RSA Factoring Challenge, which was created in 1991 to foster research in computational number theory and the practical difficulty of factoring large integers. The challenge was declared inactive in 2007. The effort behind the successful factorization of some of these numbers gives us an idea of the monumental challenges in factoring large numbers.

According to the link given in the above paragraph, there are 54 RSA numbers, ranging from 330 bits long to 2048 bits long (100 decimal digits to 617 decimal digits). Each of these numbers is a product of two prime numbers. Of these 54 numbers, 18 were successfully factored (as of the writing of this post). They were all massive efforts involving large groups of volunteers (in some cases using hundreds or thousands of computers), spanning over months or years. Some of methods used are the quadratic sieve algorithm, the number field sieve algorithm, and the general number field sieve algorithm.

The largest RSA number that was successfully factored is the RSA-768, which is 768 bits long and has 232 decimal digits (completed in December 2009). The method used was the Number Field Sieve method. There were 4 main steps in this effort. The first step is the polynomial selection, which took half a year using 80 processors. The second step is the sieving step, which took almost two years on many hundreds of machines. If only using a single core 2.2 GHz AMD Opteron processor with 2 GB RAM, the second step would take about 1500 years! The third step is the matrix step, which took a couple of weeks on a few processors. The final step took a few days, which involved a great deal of debugging.

The number field sieve method is the fastest known method for factoring large numbers that are a product of two primes (i.e. RSA moduli). The effort that went into factoring RSA-768 was massive and involved many years of complicated calculations and processing. This was only a medium size number on the list!

Another interesting observation that can be made is on the RSA numbers that have not been factored yet. There are 36 unfactored numbers in the list. One indication that RSA is secure in the current environment is that the larger numbers in the list are not yet factored (e.g. RSA-1024 which is 1024-bit long). Successful factorization of these numbers has important security implication for RSA. The largest number on the list is RSA-2048, which is 2048-bit long and has 617 digits. It is widely believed that RSA-2048 will stay unfactored in the decades to come, barring any dramatic and significant advance in computing technology.

The factoring challenge for the RSA numbers certainly provides empirical evidence that factoring is hard. Of course, no one should be complacent. We should not think that factoring will always be hard. Technology will continue to improve. A 768-bit RSA modulus was once considered secure. With the successful factorization of RSA-768, key size of 768 bits is no longer considered secure. Currently 1024 bit key size is considered secure. The RSA number RSA-1024 could very well be broken in within the next decade.

There could be new advances in factoring algorithm too. A problem that is thought to be hard may eventually turn out to be easy. Just because everyone thinks that there is no fast way of factoring, it does not mean that no such method exists. It is possible that someone has discovered such a method but decides to keep it secret in order to maintain the advantage. Beyond the issue of factoring, there could be some other vulnerabilities in RSA that can be explored and exploited by attackers.

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Fermat primality test

We now give some examples showing primality testing is a much better approach (over factoring) if the goal is to check the “prime or composite” status only. We use Fermat primality test as an example.

Example 1
Let $n=15144781$. This is a small number. So factoring would be practical as a primality test. We use it to illustrate the point that the “prime or composite” status can be determined without factoring. One option is to use Fermat’s little theorem (hence the name of Fermat primality test):

Fermat’s little theorem
If $n$ is a prime number and if $a$ is an integer that is relatively prime to $n$, then $a^{n-1} \equiv 1 \ (\text{mod} \ n)$.

Consider the contrapositive of the theorem. If we can find an $a$, relatively prime to $n$ such that $a^{n-1} \not \equiv 1 \ (\text{mod} \ n)$, then we know for sure $n$ is not prime. Such a value of $a$ is said to be a Fermat witness for the compositeness of $n$.

If a Fermat witness is found, then we can say conclusively that $n$ is composite. On the other hand, if $a$ is relatively prime to $n$ and $a^{n-1} \equiv 1 \ (\text{mod} \ n)$, then $n$ is probably a prime. We can then declare $n$ is prime or choose to run the test for a few more random values of $a$.

The exponentiation $a^{n-1} \ (\text{mod} \ n)$ is done using the fast powering algorithm, which involves a series of squarings and multiplications. Even for large moduli, the computer implementation of this algorithm is fast and efficient.

Let’s try some value of $a$, say $a=2$. Using an online calculator, we have

$2^{15144780} \equiv 1789293 \not \equiv 1 \ (\text{mod} \ 15144781)$

In this case, one congruence calculation tells us that $n=15144781$ is not prime (if it were, the congruence calculation would lead to a value of one). It turns out that $n=15144781$ is a product of two primes where $n=15144781=3733 \cdot 4057$. Of course, this is not a secure modulus for RSA. The current consensus is to use a modulus that is at least 1024-bit long.

Example 2
Let $n=15231691$. This is also a small number (in relation what is required for RSA). Once again this is an illustrative example. We calculate $a^{15231690} \ (\text{mod} \ 15231691)$ for $a=2,3,4,5,6,7$, the first few values of $a$. All such congruence values are one. We suspect that $n=15231691$ may be prime. So we randomly choose 20 values of $a$ and compute $a^{15231690} \ (\text{mod} \ 15231691)$. The following shows the results.

$\left[\begin{array}{rrr} a & \text{ } & a^{n-1} \ \text{mod} \ n \\ \text{ } & \text{ } & n=15,231,691 \\ \text{ } & \text{ } & \text{ } \\ 3,747,236 & \text{ } & 1 \\ 370,478 & \text{ } & 1 \\ 12,094,560 & \text{ } & 1 \\ 705,835 & \text{ } & 1 \\ 10,571,714 & \text{ } & 1 \\ 15,004,366 & \text{ } & 1 \\ 12,216,046 & \text{ } & 1 \\ 10,708,300 & \text{ } & 1 \\ 6,243,738 & \text{ } & 1 \\ 1,523,626 & \text{ } & 1 \\ 10,496,554 & \text{ } & 1 \\ 10,332,033 & \text{ } & 1 \\ 10,233,123 & \text{ } & 1 \\ 3,996,691 & \text{ } & 1 \\ 4,221,958 & \text{ } & 1 \\ 3,139,943 & \text{ } & 1 \\ 1,736,767 & \text{ } & 1 \\ 12,672,150 & \text{ } & 1 \\ 12,028,143 & \text{ } & 1 \\ 8,528,642 & \text{ } & 1 \end{array}\right]$

For all 20 random values of $a$, $a^{15231690} \equiv 1 \ (\text{mod} \ 15231691)$. This represents strong evidence (though not absolute proof) that $n=15231691$ is a prime. In fact, we can attach the following probability statement to the above table of 20 random values of $a$.

If $n=15231691$ were a composite number that has at least one Fermat witness, there is at most a 0.0000953674% chance that 20 randomly selected values of $a$ are not Fermat witnesses.

In other words, if $n=15231691$ were a composite number that has at least one Fermat witness, there is at most a 0.0000953674% chance of getting 20 1’s in the above computation.

In general, if $n$ has at least one Fermat witness, the probability that all $k$ randomly selected values of $a$ with $1 are not Fermat witnesses is at most $0.5^k$. For $k=20$, $0.5^{20}=0.000000953674$, which is 0.0000953674%. The probability statement should give us enough confidence to consider $n=15231691$ a prime number.

There is a caveat that has to be mentioned. For the above probability statement to be valid, the number $n$ must have at least one Fermat witness. If a number $n$ is composite, we would like the test to produce a Fermat witness. It turns out that there are composite numbers that have no Fermat witnesses. These numbers are called Carmichael numbers. If $n$ is such a number, $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ for any $a$ that is relatively prime to the number $n$. In other words, the Fermat test will always indicate “probably prime” for Carmichael numbers. Unless you are lucky and randomly pick a value of $a$ that shares a common prime factor with $n$, the Fermat test will always incorrectly identify a Carmichael number $n$ as prime. Fortunately Carmichael numbers are rare, even though there are infinitely many of them. In this previous post, we estimate that a randomly selected 1024-bit odd integer has a less than one in $10^{88}$ chance of being a Carmichael number!

The Fermat test is a powerful test when the number being tested is a prime number or a composite number that has a Fermat witness. For Carmichael numbers, the test is likely to produce a false positive (identifying a composite number as prime). Thus the existence of Carmichael numbers is the biggest weakness of the Fermat test. Fortunately Carmichael numbers are rare. Though they are rare, their existence may still make the Fermat test unsuitable in some situation, e.g., when you test a number provided by your adversary. If you really want to avoid situations like these, you can always switch to the Miller-Rabin test.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Congruence Arithmetic and Fast Powering Algorithm

In some cryptography applications such as RSA algorithm, it is necessary to compute $\displaystyle a^w$ modulo $m$ where the power $w$ and the modulus $m$ are very large numbers. We discuss and demonstrate an efficient algorithm that can handle such calculations. This general algorithm has various names such as fast powering algorithm, square-and-multiply algorithm and exponentiation by squaring.

The problem at hand is to compute $a^w \ (\text{mod} \ m)$. The naïve approach is to compute by repeatedly multiplying by $a$ and reducing modulo $m$. When the power $w$ is large (e.g. an integer with hundreds of digits), this approach is difficult or even impossible (given the current technology). In this post we discuss an alternative that is known as the fast powering algorithm.

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An Example

Compute $1286^{1171}$ modulo $1363$.

Using the naïve approach described earlier, we would do something like the following:

\displaystyle \begin{aligned} 1286^{1171} &=(1286 \cdot 1286) \cdot 1286^{1269} \equiv 477 \cdot 1286^{1269} \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \\&=(477 \cdot 1286) \cdot 1286^{1268} \equiv 72 \cdot 1286^{1268} \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \\&=(72 \cdot 1286) \cdot 1286^{1267} \equiv 1271 \cdot 1286^{1267} \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \\&\text{ }\cdots \\&\text{ }\cdots \\&\text{ }\cdots \end{aligned}

In this naïve approach, we would multiply two numbers at a time and then reduce the result modulo $1363$ so that the numbers do not get too large. The above example would involve $1170$ multiplications and then $1170$ divisions for the reduction modulo $1363$. Great difficulty comes when the power is not $1171$ and instead is an integer with hundreds or even thousands of digits.

Note that in the above naïve approach, the power is reduced by one at each step. In the fast power alternative, the power comes down by an exponent of two in each step. The idea is to use the binary expansions of the exponent $1171$ to transform the computation of $1286^{1171}$ into a series of squarings and multiplications. To this end, we write $1171$ as a sum of powers of two as follows:

$(1) \ \ \ \ \ \ \ \ \ 1171=2^0+2^1+2^4+2^7+2^{10}$

Next we compute $1286^{2^0},1286^{2^1},1286^{2^2},1286^{2^3},\cdots,1286^{2^{10}}$ modulo $1363$. Note that each term is the square of the preceding term, hence the word square in the name “square-and-multiply”. To make it easier to see, we put the results in the following table.

$\displaystyle (2) \ \ \ \ \ \ \ \ \ \begin{bmatrix} \text{ i }&\text{ }&1286^{2^i}&\text{ }&\text{squaring}&\text{ }&\text{modulo } 1363&\text{ }&\text{ } \\\text{ }&\text{ }&\text{ } \\ 0&\text{ }&1286^{2^0}&\text{ }&\text{ }&\text{ }&\equiv 1286&\text{ }&* \\ 1&\text{ }&1286^{2^1}&\text{ }&\equiv 1286^2&\text{ }&\equiv 477&\text{ }&* \\ 2&\text{ }&1286^{2^2}&\text{ }&\equiv 477^2&\text{ }&\equiv 1271&\text{ }&\text{ } \\ 3&\text{ }&1286^{2^3}&\text{ }&\equiv 1271^2&\text{ }&\equiv 286&\text{ }&\text{ } \\ 4&\text{ }&1286^{2^4}&\text{ }&\equiv 286^2&\text{ }&\equiv 16&\text{ }&* \\ 5&\text{ }&1286^{2^5}&\text{ }&\equiv 16^2&\text{ }&\equiv 256&\text{ }&\text{ } \\ 6&\text{ }&1286^{2^6}&\text{ }&\equiv 256^2&\text{ }&\equiv 112&\text{ }&\text{ } \\ 7&\text{ }&1286^{2^7}&\text{ }&\equiv 112^2&\text{ }&\equiv 277&\text{ }&* \\ 8&\text{ }&1286^{2^8}&\text{ }&\equiv 277^2&\text{ }&\equiv 401&\text{ }&\text{ } \\ 9&\text{ }&1286^{2^9}&\text{ }&\equiv 401^2&\text{ }&\equiv 1330&\text{ }&\text{ } \\ 10&\text{ }&1286^{2^{10}}&\text{ }&\equiv 1330^2&\text{ }&\equiv 1089&\text{ }&* \end{bmatrix}$

Note that the rows marked by * in the above table are the results that we need. In the above table, there are 10 multiplications for the squarings and 10 divisions for the reduction modulo $1363$.

Now $1286^{1171}$ is calculated as follows:

\displaystyle \begin{aligned} (3) \ \ \ \ \ \ \ \ \ 1286^{1171} &=1286^{2^0} \cdot 1286^{2^1} \cdot 1286^{2^4} \cdot1286^{2^7} \cdot 1286^{2^{10}} \\&\equiv 1286 \ \ \cdot 477 \ \ \ \ \cdot 16 \ \ \ \ \ \cdot 277 \ \ \ \ \cdot 1089 \ \ \text{mod} \ 1363 \\&\equiv 72 \ \ \ \ \cdot 16 \ \ \ \ \ \cdot 277 \ \ \ \ \cdot 1089 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \\&\equiv 1152 \ \ \cdot 277 \ \ \cdot 1089 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \\&\equiv 162 \ \ \cdot 1089 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \\&\equiv 591 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 1363 \end{aligned}

We have the answer $1286^{1171} \equiv 591 \ (\text{mod} \ 1363)$. The calculation in (3) is the step that gives the word “multiply” in the name “square-and-multiply”. In this step, we multiply the results obtained from the previous step.

We now tally up the amount of work that is done. The calculation in table (2) requires 10 multiplications for the squaring and 10 divisions for the reduction modulo $1363$. The calculation in (3) requires 4 multiplications and 4 divisions for the reduction modulo $1363$. Together, there are 14 multiplications and 14 divisions. In contrast, the naïve approach would require 1170 multiplications and 1170 divisions!

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Description of Fast Powering Algorithm

To compute $a^w \equiv (\text{mod} \ m)$, use the following steps. The following steps correspond with the steps in the above example.

Step (1)

Compute the binary expansions of the power $w$.

$\displaystyle w=C_0 +C_1 \cdot 2^{1}+C_2 \cdot 2^{2}+\cdots+C_{k-1} \cdot 2^{k-1}+C_k \cdot 2^{k}$

where each $j$, $C_j=0$ or $C_j=1$. In particular, we assume that $C_k=1$.

Step (2)

For each $j=0,1,2,\cdots,k$, compute $\displaystyle a^{2^j} \equiv A_j$ modulo $m$. Note that each $\displaystyle a^{2^j} \equiv A_j$ is the result of squaring the previous term $\displaystyle a^{2^{j-1}} \equiv A_{j-1}$. We arrange the calculation in the following table.

$\displaystyle (2) \ \ \ \ \ \ \ \ \ \begin{bmatrix} \text{ i }&\text{ }&a^{2^i}&\text{ }&\text{squaring}&\text{ }&\text{modulo } m&\text{ }&\text{ } \\\text{ }&\text{ }&\text{ } \\ 0&\text{ }&a^{2^0}&\text{ }&\text{ }&\text{ }&A_0\equiv a&\text{ }&\text{ } \\ 1&\text{ }&a^{2^1}&\text{ }&\equiv A_0^2&\text{ }&\equiv A_1&\text{ }&\text{ } \\ 2&\text{ }&a^{2^2}&\text{ }&\equiv A_1^2&\text{ }&\equiv A_2&\text{ }&\text{ } \\ 3&\text{ }&a^{2^3}&\text{ }&\equiv A_2^2&\text{ }&\equiv A_3&\text{ }&\text{ } \\ \text{ }&\text{ }&\cdots&\text{ }&\cdots&\text{ }&\cdots&\text{ }&\text{ } \\ \text{ }&\text{ }&\cdots&\text{ }&\cdots&\text{ }&\cdots&\text{ }&\text{ } \\ \text{ }&\text{ }&\cdots&\text{ }&\cdots&\text{ }&\cdots&\text{ }&\text{ } \\ k-1&\text{ }&a^{2^{k-1}}&\text{ }&\equiv A_{k-2}^2&\text{ }&\equiv A_{k-1}&\text{ }&\text{ } \\ k&\text{ }&a^{2^k}&\text{ }&\equiv A_{k-1}^2&\text{ }&\equiv A_k&\text{ }&\text{ } \end{bmatrix}$

Step (3)

Compute $a^w \equiv (\text{mod} \ m)$ using the following derivation.

\displaystyle \begin{aligned}(3) \ \ \ \ \ \ \ \ \ a^{w}&=a^{C_0 +C_1 \cdot 2^{1}+C_2 \cdot 2^{2}+\cdots+C_{k-1} \cdot 2^{k-1}+C_k \cdot 2^{k}} \\&=a^{C_0} \cdot a^{C_1 \cdot 2^{1}} \cdot a^{C_2 \cdot 2^{2}} \cdots a^{C_{k-1} \cdot 2^{k-1}} \cdot a^{C_k \cdot 2^{k}} \\&=a^{C_0} \cdot (a^{2^{1}})^{C_1} \cdot (a^{2^{2}})^{C_2} \cdots (a^{2^{k-1}})^{C_{k-1}} \cdot (a^{2^{k}})^{C_k} \\&\equiv A_0^{C_0} \cdot (A_1)^{C_1} \cdot (A_2)^{C_2} \cdots (A_{k-1})^{C_{k-1}} \cdot (A_k)^{C_k} \ \ \ \ \ (\text{mod} \ m) \end{aligned}

The last line in (3) is to be further reduced modulo $m$. In the actual calculation, only the terms with $C_j=1$ need to be used.

We now establish an upper bound for the number multiplications. Step (2) requires $k$ multiplications and $k$ divisions to reduce modulo $m$. Step (3) requires at most $k$ many multiplications since some of the $C_j$ many be zero. Step (3) also requires at most $k$ many divisions to reduce modulo $m$. So altogether, the algorithm requires at most $2k$ multiplications and $2k$ divisions.

From Step (1), we know that $\displaystyle 2^k \le w$. Take natural log of both sides, we have $\displaystyle k \le \frac{\text{ln}(w)}{\text{ln}(2)}$ and $\displaystyle 2 \cdot k \le \frac{2 \cdot \text{ln}(w)}{\text{ln}(2)}$. So the fast powering algorithm requires at most

$\displaystyle \frac{2 \cdot \text{ln}(w)}{\text{ln}(2)}$

many multiplications and at most that many divisions to compute the congruence calculation $a^w \equiv (\text{mod} \ m)$.

For example, when the power $w=2^{127}-1$, which is a Mersenne prime, which has 39 digits. Now $w \approx 2^{127}$. By the above calculation, the fast powering algorithm would take at most 254 multiplications and at most 254 divisions to do the power congruence computation.

The fast powering calculations demonstrated in this post can be done by hand (using a hand-held calculator). In real applications, such calculations should of course be done in a computer.

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Another Example

Use the fast power algorithm to show that

$4030^{2657} \equiv 21144 \ (\text{mod} \ 55049)$

$21144^{79081} \equiv 4030 \ (\text{mod} \ 55049)$

Note that one congruence is encryption and the other one is decryption. We demonstrate the second calculation.

In doing the second calculation, we use a little bit of help from Fermat’s little theorem. The modulus $55049$ is a prime number. So $21144^{55048} \equiv 1 \ (\text{mod} \ 55049)$. Thus we have:

$21144^{79081}=21144^{24033} \cdot 21144^{55048} \equiv 21144^{24033} \ (\text{mod} \ 55049)$

Step (1)

The binary expansions of $24033$ are:

$24033=2^0+2^5+2^6+2^7+2^8+2^{10}+2^{11}+2^{12}+2^{14}$

Step (2)

Compute $21144^{2^j}$ modulo $55049$ for each $j$. The computation is displayed in the following table. The rows with * are the results that we need for Step (3).

$\displaystyle \begin{bmatrix} \text{ i }&\text{ }&21144^{2^i}&\text{ }&\text{squaring}&\text{ }&\text{modulo } 55049&\text{ }&\text{ } \\\text{ }&\text{ }&\text{ } \\ 0&\text{ }&21144^{2^0}&\text{ }&\text{ }&\text{ }&\equiv 21144&\text{ }&* \\ 1&\text{ }&21144^{2^1}&\text{ }&\equiv 21144^2&\text{ }&\equiv 15807&\text{ }&\text{ } \\ 2&\text{ }&21144^{2^2}&\text{ }&\equiv 15807^2&\text{ }&\equiv 48887&\text{ }&\text{ } \\ 3&\text{ }&21144^{2^3}&\text{ }&\equiv 48887^2&\text{ }&\equiv 41483&\text{ }&\text{ } \\ 4&\text{ }&21144^{2^4}&\text{ }&\equiv 41483^2&\text{ }&\equiv 7549&\text{ }&\text{ } \\ 5&\text{ }&21144^{2^5}&\text{ }&\equiv 7549^2&\text{ }&\equiv 11686&\text{ }&* \\ 6&\text{ }&21144^{2^6}&\text{ }&\equiv 11686^2&\text{ }&\equiv 41076&\text{ }&* \\ 7&\text{ }&21144^{2^7}&\text{ }&\equiv 41076^2&\text{ }&\equiv 40975&\text{ }&* \\ 8&\text{ }&21144^{2^8}&\text{ }&\equiv 40975^2&\text{ }&\equiv 11174&\text{ }&* \\ 9&\text{ }&21144^{2^9}&\text{ }&\equiv 11174^2&\text{ }&\equiv 7144&\text{ }&\text{ } \\ 10&\text{ }&21144^{2^{10}}&\text{ }&\equiv 7144^2&\text{ }&\equiv 6313&\text{ }&* \\ 11&\text{ }&21144^{2^{11}}&\text{ }&\equiv 6313^2&\text{ }&\equiv 53542&\text{ }&* \\ 12&\text{ }&21144^{2^{12}}&\text{ }&\equiv 53542^2&\text{ }&\equiv 14040&\text{ }&* \\ 13&\text{ }&21144^{2^{13}}&\text{ }&\equiv 14040^2&\text{ }&\equiv 46180&\text{ }&\text{ } \\ 14&\text{ }&21144^{2^{14}}&\text{ }&\equiv 46180^2&\text{ }&\equiv 49189&\text{ }&* \end{bmatrix}$

Step (3)

Compute $21144^{79081} \ (\text{mod} \ 55049)$.

\displaystyle \begin{aligned} 21144^{24033} &=21144^{2^0} \cdot 21144^{2^5} \cdot 21144^{2^6} \cdot 21144^{2^7} \cdot 21144^{2^{8}} \cdot 21144^{2^{10}} \cdot 21144^{2^{11}} \cdot 21144^{2^{12}} \cdot 21144^{2^{14}} \\&\equiv 21144 \cdot 11686 \cdot 41076 \cdot 40975 \cdot 11174 \cdot 6313 \cdot 53542 \cdot 14040 \cdot 49189 \ \ \text{mod} \ 55049 \\&\equiv 25665 \cdot 40975 \cdot 11174 \cdot 6313 \cdot 53542 \cdot 14040 \cdot 49189 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \\&\equiv 22328 \cdot 11174 \cdot 6313 \cdot 53542 \cdot 14040 \cdot 49189 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \\&\equiv 11004 \cdot 6313 \cdot 53542 \cdot 14040 \cdot 49189 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \\&\equiv 51463 \cdot 53542 \cdot 14040 \cdot 49189 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \\&\equiv 9300 \cdot 14040 \cdot 49189 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \\&\equiv 50821 \cdot 49189 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \\&\equiv 4030 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{mod} \ 55049 \end{aligned}

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$\copyright \ 2013 \text{ by Dan Ma}$

# How to toss a coin

In this post, we demonstrate how to simulate a random coin toss using a cryptographic algorithm that was proposed by Rivest, Shamir and Adlemen in 1982 (the creators of the RSA algorithm). Tossing a coin using a cryptographic algorithm is an example of a game of mental poker.

The term mental poker refers to the game of poker played over long distance that has a mechanism for ensuring a fair game without the need for a trusted third party. Mental poker can also refer to other cryptographic games played over long distance without the need for a trusted third party (e.g. tossing a coin over long distance).

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Setting the Algorithm

A full discussion of the algorithm used here can be found in the previous post Fermat’s Little Theorem and Mental Poker.

Andy and Becky are in different locations and they would like to simulate a random coin toss. First they need to agree on two positive integers that are to represent head and tail, say $m_h$ for head and $m_t$ for tail. These two numbers should be less than the prime number $p$ discussed in the next paragraph.

They now need to encrypt the numbers before tossing the coin. Both players agree on a large prime number $p$. If the goal of preventing or minimizing cheating is important, the players should choose a prime number that is as large as feasible computationally speaking.

With the prime number $p$ decided, each of the players chooses an encryption-decryption key, which is a pair of positive integers $x_0$ and $x_1$ such that

$x_0 \cdot x_1 \equiv 1 \ (\text{mod} \ p-1)$,

meaning that $x_0 \cdot x_1=1+(p-1) \cdot k$ for some integer $k$.

Each of the players chooses such a pair of numbers. In terms of notation, Andy chooses $a_0$ and $a_1$. Becky chooses $b_0$ and $b_1$. Each player chooses the number pair independently and keeps it secret from the other player.

The number with the $0$-subscript is the encryption key and the number with the $1$-subscript is the decryption key. The following are the encryption functions, expressed in congruence modulo $p$, for Andy and Becky, respectively.

$f_a(m) \equiv m^{a_0} \ (\text{mod} \ p) \ \ \ \ \ \ \ \text{Andy}$

$f_b(m) \equiv m^{b_0} \ (\text{mod} \ p) \ \ \ \ \ \ \ \text{Becky}$

For example, if Andy wants to encrypt the number $m$, he raises $m$ to the power of $a_0$ and looks for the remainder upon division by $p$. The remainder will be denoted by $f_a(m)$. The function $f_b(m)$ works similarly for Becky.

The following are the decryption functions, expressed in congruence modulo $p$, for Andy and Becky, respectively.

$g_a(c) \equiv c^{a_1} \ (\text{mod} \ p) \ \ \ \ \ \ \ \text{Andy}$

$g_b(c) \equiv c^{b_1} \ (\text{mod} \ p) \ \ \ \ \ \ \ \text{Becky}$

For example, if Andy wants to decrypt the number $c=f_a(m)$, he raises $c$ to the power of $a_1$ and looks for the remainder upon division by $p$. The remainder will be denoted by $g_a(m)$, which will be the original number $m$. The proof of this fact is based on the Fermat’s Little Theorem (see the previous post Fermat’s Little Theorem and Mental Poker).

The function $g_b(m)$ works similarly for Becky.

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A Numerical Example

For illustration, we use a small prime number. We use $p=7919$. We emphasize that this is just for illustration. In an application where the chance for cheating is to be minimized, a large prime number should be used.

Andy and Becky use the following assignment for Head and Tail.

$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Head}&\text{ }&\text{Tail} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Original}&\text{ }&\text{2,500}&\text{ }&\text{5,000} \end{bmatrix}$

For this illustration, Andy chooses $a_0=\text{47}$ and $a_1=\text{52,899}$ by letting $k=\text{314}$ in the equation below. Becky chooses $b_0=\text{71}$ and $b_1=\text{26,319}$ by letting $k=\text{236}$ in the following equation.

$x_0 \cdot x_1=1+7918 \cdot k$

Andy and Becky choose these keys independently and they keep them secret without letting the other person know. Andy encrypts both the head and tail numbers as follows:

$2500^{47} \equiv 7518=f_a(2500) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

$5000^{47} \equiv 698=f_a(5000) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

The encrypted numbers and the original numbers are displayed in the following table.

$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Head}&\text{ }&\text{Tail} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Original}&\text{ }&\text{2,500}&\text{ }&\text{5,000} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Encrypted by Andy}&\text{ }&\text{7,518}&\text{ }&\text{698} \end{bmatrix}$

Of course, the above table is kept secret from Becky. However, the encrypted numbers (just the encrypted numbers) are sent to Becky for random selection.

Becky selects one of the encrypted numbers for herself (perhaps thru a coin toss). Then the other encrypted number is the choice for Andy. Suppose Becky selects $7518$. Becky then encrypted the number $7518$ using her key.

$7518^{71} \equiv 1341=f_b(7518) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

Becky passes the encrypted number $1341$ (her selection) and $698$ (Andy’s selection) back to Andy. The following table lists out the numbers received by Andy.

$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Andy}&\text{ }&\text{Becky} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Becky's selection}&\text{ }&\text{698}&\text{ }&\text{7,518} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Encrypted by Becky}&\text{ }&\text{ }&\text{ }&\text{1,341} \end{bmatrix}$

Andy decrypts his number $698$ and gets back $2500$, which is tail. He also decrypts $1341$ and obtains $223$, which he passes back to Becky.

$698^{52899} \equiv 5000=g_a(698) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

$1341^{52899} \equiv 223=g_a(1341) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$

The following summarizes the results up to this point.

$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Andy}&\text{ }&\text{Becky} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Becky's selection}&\text{ }&\text{698}&\text{ }&\text{7,518} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Encrypted by Becky}&\text{ }&\text{ }&\text{ }&\text{1,341} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Decrypted by Andy}&\text{ }&\text{5,000}&\text{ }&\text{223} \end{bmatrix}$

Once Becky gets the decrypted number $223$ from Andy, she decrypts it using her own key to obtain $2500$, which is a head.

$223^{26319} \equiv 2500=g_b(223) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)$

The following table summarizes the results of the coin toss.

$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Andy}&\text{ }&\text{Becky} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Becky's selection}&\text{ }&\text{698}&\text{ }&\text{7,518} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Encrypted by Becky}&\text{ }&\text{ }&\text{ }&\text{1,341} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Decrypted by Andy}&\text{ }&\text{5,000}&\text{ }&\text{223} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Decrypted by Becky}&\text{ }&\text{ }&\text{ }&\text{2,500} \\ \text{ }&\text{ }&\text{ }&\text{ }&\text{ } \\ \text{Results of Coin Toss}&\text{ }&\text{5,000}&\text{ }&\text{2,500} \end{bmatrix}$

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Numerical Calculation

We now show some of the calculations involved in the above encryption and decryption. We show three calculations.

$5000^{47} \equiv 698=f_a(5000) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

$698^{52899} \equiv 5000=g_a(698) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

$223^{26319} \equiv 2500=g_b(223) \ (\text{mod} \ 7919) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)$

Even with the small prime number $p=7919$, the calculation is not done directly. For example, unless special software is used, $f_a(5000)$ is not found by calculating $5000^{47}$ and then finding the remainder upon division by $7919$. The following demonstrates a “divide and conquer” approach for the calculation for $(2)$ where in each step the exponent is reduced by half.

\displaystyle \begin{aligned} 5000^{47}&\equiv (5000^2)^{23} \cdot 5000 \ (\text{mod} \ 7919) \\&\text{ } \\&=7636^{23} \cdot 5000 \equiv (7636^2)^{11} \cdot 7636 \cdot 5000 \\&\text{ } \\&\equiv 899^{11} \cdot 7636 \cdot 5000 \\&\text{ } \\&\equiv 899^{11} \cdot 2501 \equiv (899^2)^5 \cdot 899 \cdot 2501 \\&\text{ } \\&\equiv 463^5 \cdot 899 \cdot 2501 \\&\text{ } \\&\equiv 463^5 \cdot 7322 \equiv (463^2)^2 \cdot 463 \cdot 7322 \\&\text{ } \\&\equiv 556^2 \cdot 463 \cdot 7322 \\&\text{ } \\&\equiv 556^2 \cdot 754 \\&\text{ } \\&\equiv 295 \cdot 754 \\&\text{ } \\&\equiv 698 \ (\text{mod} \ 7919) \end{aligned}

In each step in the above calculation, we use the division algorithm to obtain the remainder. For example, to go from the first line to the second line, divide $5000^2$ by $7919$ to obtain the remainder of $7636$. The number $698$ in the last step is the remainder when $295 \cdot 754$ is divided by $7919$. These calculations are tedious if done by hand and should be done by computer.

The calculation for $(4)$ is that $698^{52899} \equiv 5000 \ (\text{mod} \ 7919)$. In other word, decrypting the encrypted number of $698$ recovers the original number of $5000$. This calculation can be shortened by using Fermat’s Little Theorem, which implies that $698^{7919-1} \equiv 1 \ (\text{mod} \ 7919)$. Thus we have:

$698^{47508} \equiv 698^{6 \cdot 7918} \equiv 1 \ (\text{mod} \ 7919)$

So we can reduce $47508$ from the exponent $52899$, leaving the exponent $5391$. We have:

$698^{52899} \equiv 698^{47508} \cdot 698^{5391} \equiv 1 \cdot 698^{5391} \ (\text{mod} \ 7919)$

The following uses the “divide and conquer” approach to compute $698^{5391}$ modulo $7919$.

\displaystyle \begin{aligned} 698^{5391}&\equiv (698^2)^{2695} \cdot 698 \ (\text{mod} \ 7919) \\&\text{ } \\&=4145^{2695} \cdot 698 \equiv (4145^2)^{1347} \cdot 4145 \cdot 698 \\&\text{ } \\&\equiv 4714^{1347} \cdot 4145 \cdot 698 \\&\text{ } \\&\equiv 4714^{1347} \cdot 2775 \equiv (4714^2)^{673} \cdot 4714 \cdot 2775 \\&\text{ } \\&\equiv 1082^{673} \cdot 4714 \cdot 2775 \\&\text{ } \\&\equiv 1082^{673} \cdot 7081 \equiv (1082^2)^{336} \cdot 1082 \cdot 7081 \\&\text{ } \\&\equiv 6631^{336} \cdot 1082 \cdot 7081 \\&\text{ } \\&\equiv 6631^{336} \cdot 3969 \equiv (6631^2)^{168} \cdot 3969 \\&\text{ } \\&\equiv 3873^{168} \cdot 3969 \equiv (3873^2)^{84} \cdot 3969 \\&\text{ } \\&\equiv 1543^{84} \cdot 3969 \equiv (1543^2)^{42} \cdot 3969 \\&\text{ } \\&\equiv 5149^{42} \cdot 3969 \equiv (5149^2)^{21} \cdot 3969\\&\text{ } \\&\equiv 7308^{21} \cdot 3969 \equiv (7308^2)^{10} \cdot 7308 \cdot 3969 \\&\text{ } \\&\equiv 1128^{10} \cdot 7308 \cdot 3969 \\&\text{ } \\&\equiv 1128^{10} \cdot 6074 \equiv (1128^2)^{5} \cdot 6074\\&\text{ } \\&\equiv 5344^5 \cdot 6074 \equiv (5344^2)^2 \cdot 5344 \cdot 6074 \\&\text{ } \\&\equiv 2422^2 \cdot 5344 \cdot 6074 \\&\text{ } \\&\equiv2422^2 \cdot 7394 \\&\text{ } \\&\equiv 6024 \cdot 7394 \\&\text{ } \\&\equiv 5000 \ (\text{mod} \ 7919) \end{aligned}

The calculation for $(6)$ is $223^{26319} \equiv 2500 \ (\text{mod} \ 7919)$. We can also get an assist from the Fermat’s Little Theorem. In this particular case, $223^{7918} \equiv 1 \ (\text{mod} \ 7919)$. With $26319=3 \cdot 7918+2565$, we only need to calculate $223^{2565} \ (\text{mod} \ 7919)$, which is done below.

\displaystyle \begin{aligned} 223^{2565}&\equiv (223^2)^{1282} \cdot 223 \ (\text{mod} \ 7919) \\&\text{ } \\&=2215^{1282} \cdot 223 \equiv (2215^2)^{641} \cdot 223 \\&\text{ } \\&\equiv 4364^{641} \cdot 223 \equiv (4364^2)^{320} \cdot 4364 \cdot 223 \\&\text{ } \\&\equiv 7220^{320} \cdot 4364 \cdot 223 \\&\text{ } \\&\equiv 7220^{320} \cdot 7054 \equiv (7220^2)^{160} \cdot 7054 \\&\text{ } \\&\equiv 5542^{160} \cdot 7054 \equiv (5442^2)^{80} \cdot 7054 \\&\text{ } \\&\equiv 3882^{80} \cdot 7054 \equiv (3882^2)^{40} \cdot 7054 \\&\text{ } \\&\equiv 67^{40} \cdot 7054 \equiv (67^2)^{20} \cdot 7054 \\&\text{ } \\&\equiv 4489^{20} \cdot 7054 \equiv (4489^2)^{10} \cdot 7054\\&\text{ } \\&\equiv 5185^{10} \cdot 7054 \equiv (5185^2)^{5} \cdot 7054 \\&\text{ } \\&\equiv 7139^{5} \cdot 7054 \equiv (7139^2)^{2} \cdot 7139 \cdot 7054 \\&\text{ } \\&\equiv 6556^2 \cdot 7139 \cdot 7054 \\&\text{ } \\&\equiv 6556^2 \cdot 1585 \\&\text{ } \\&\equiv 4723 \cdot 1585 \\&\text{ } \\&\equiv 2500 \ (\text{mod} \ 7919) \end{aligned}

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$\copyright \ 2013 \text{ by Dan Ma}$

# Fermat’s Little Theorem and Mental Poker

In this post we demonstrate another use of the Fermat’s Little Theorem.

How can two people play poker when they are not sitting face to face from each other? If the game of poker is played over long distance (e.g. via a telephone line or some electronic communication channel), there will be a need to ensure a fair game. For example, the two players must use the same deck of cards (ensuring that there will be no duplicates). The deck of cards will need to be well shuffled. Each player cannot see the cards of the other player. One solution is to use a trusted third party to do the shuffling and selecting of cards. If a third party cannot be found or it is felt that the third party cannot be trusted to be fair, then one should consider the cryptographic solution described in this post. This soultion was proposed by Rivest, Shamir and Adlemen in 1982 (the creators of the RSA algorithm).

The term mental poker refers to the game of poker played over long distance that has a mechanism for ensuring a fair game without the need for a trusted third party. Mental poker can also refer to other cryptographic games played over long distance without the need for a trusted third party (e.g. tossing a coin over long distance).

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Setting Up the Deck of Cards

Let’s say that the players are Andy and Becky. Since they are not using a physical deck of cards, they need to represent the cards by numbers. Let’s say that they agree to number the cards as follows:

$\displaystyle \heartsuit 2=1020 \ \ \ \ \ \ \ \ \ \ \ \diamondsuit 2=2020 \ \ \ \ \ \ \ \ \ \ \ \spadesuit 2=3020 \ \ \ \ \ \ \ \ \ \ \ \clubsuit 2=4020$

$\displaystyle \heartsuit 3=1030 \ \ \ \ \ \ \ \ \ \ \ \diamondsuit 3=2030 \ \ \ \ \ \ \ \ \ \ \ \spadesuit 3=3030 \ \ \ \ \ \ \ \ \ \ \ \clubsuit 3=4030$

$\displaystyle \heartsuit 4=1040 \ \ \ \ \ \ \ \ \ \ \ \diamondsuit 4=2040 \ \ \ \ \ \ \ \ \ \ \ \spadesuit 4=3040 \ \ \ \ \ \ \ \ \ \ \ \clubsuit 4=4040$

$\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots$

$\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots$

$\cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdots$

$\displaystyle \heartsuit Q=1120 \ \ \ \ \ \ \ \ \ \ \ \diamondsuit Q=2120 \ \ \ \ \ \ \ \ \ \ \spadesuit Q=3120 \ \ \ \ \ \ \ \ \clubsuit Q=4120$

$\displaystyle \heartsuit K=1130 \ \ \ \ \ \ \ \ \ \ \ \diamondsuit K=2130 \ \ \ \ \ \ \ \ \ \spadesuit K=3130 \ \ \ \ \ \ \ \ \clubsuit K=4130$

$\displaystyle \heartsuit A=1140 \ \ \ \ \ \ \ \ \ \ \ \diamondsuit A=2140 \ \ \ \ \ \ \ \ \ \ \spadesuit A=3140 \ \ \ \ \ \ \ \ \ \clubsuit A=4140$

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The card numbers need to be encrypted before they can be passed between the two players. Here’s how it works.

Both players agree to choose a large prime number $p$. This number $p$ needs to be larger than all the card numbers and the encrypted card numbers. The larger $p$ is, the harder it will be for any one of the players to cheat.

Now each of the players needs to choose an encryption-decryption key (a padlock) that the player keeps secret. Let’s start with Andy. He chooses a pair of positive numbers $a_0$ and $a_1$ such that the following holds:

$a_0 \cdot a_1 \equiv 1 \ (\text{mod} \ p-1)$

Equivalently the pair $a_0$ and $a_1$ satisfies the equation $a_0 \cdot a_1=1+(p-1) \cdot k$ for some integer $k$. The number $a_0$ will be used for locking (encryption) and the number $a_1$ will be used for unlocking (decryption). Andy will also keep this pair of numbers away from Becky.

How will Andy use this padlock? Suppose that $m$ is a number to be encrypted. To encrypt the number, Andy raises $m$ to the power of $a_0$ and then finds the remainder upon division by $p$. He will call the remainder $f_a(m)$. Using congruence notation, the following is the encryption function:

$f_a(m) \equiv m^{a_0} \ (\text{mod} \ p)$

If Andy needs to recover $m$ from the encrypted card number $c=f_a(m)$, all he has to do is to raise $c$ to the power of $a_1$ and then find the remainder upon division by $p$. Call the remainder $g_a(c)$, which will be the original card number $m$. Using congruence notation, the following is the decryption function:

$g_a(c) \equiv c^{a_1} \ (\text{mod} \ p)$

The decrypted number is the original number. Thus we have $g_a(c)=m$. A proof of this relies on the Fermat’s Little Theorem (see proof).

Because the numbers involved are usually large, no one will try to raise $m$ to the power of $a_0$ and then divides by $p$ to find the remainder. Instead, Andy should use special software. If software is not available, Andy can rely on congruence modulo arithmetic, which should also be done by a computer. See below for a demonstration of the congruence modulo arithmetic.

The other player Becky also needs a padlock. Specifically, she chooses a pair of numbers $b_0$ and $b_1$ that satisfy the following:

$b_0 \cdot b_1 \equiv 1 \ (\text{mod} \ p-1)$

This pair of number serves the same purpose as the pair that belongs to Andy. Of course, $b_0$ and $b_1$ need to be kept secret from Andy. The following shows the encryption and decryption functions for Becky’s padlock.

$f_b(m) \equiv m^{b_0} \ (\text{mod} \ p)$

$g_b(c) \equiv c^{b_1} \ (\text{mod} \ p)$

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How to Play the Game

Suppose the card numbers are $m_1, m_2, m_3, \cdots, m_{52}$ (the above is one example of card number assigment). Andy then encrypts the card number using his encryption function $f_a(m)$. The following lists the encrypted card numbers.

$\displaystyle f_a(m_1),\ f_a(m_2), \ f_a(m_3),\cdots,f_a(m_{52})$

Andy then passes these encrypted card numbers to Becky. She shuffles the encrypted deck thorughly. She then chooses a 5-card hand for Andy. Becky then chooses another 5-card hand for herself. Becky uses her key to encrypt her 5-card hand. Becky passes both 5-card hands to Andy. The following shows what Becky passes to Andy.

Andy’s 5-card hand: $f_a(m_i) \equiv m_i^{a_0} \ (\text{mod} \ p)$ for 5 distinct values of $i$.

Becky’s 5-card hand: $f_b(f_a(m_j)) \equiv f_a(m_j)^{b_0} \ (\text{mod} \ p)$ for 5 distinct values of $j$.

Once Andy gets the two 5-card hands, he decrypts his own 5-card hand and gets back the original card numbers. He also decrypts Becky’s 5-card hand and passes that back to Becky.

Andy’s 5-card hand: $g_a(f_a(m_i)) \equiv (m_i^{a_0})^{a_1} \equiv m_i \ (\text{mod} \ p)$

Becky’s 5-card hand: $g_a(f_b(f_a(m_j))) \equiv (f_a(m_j)^{b_0})^{a_1}=(f_a(m_j)^{a_1})^{b_0} \equiv m_j^{b_0} \ (\text{mod} \ p)$, which Andy passes back to Becky.

Once Becky’s recieves her 5-card hand back from Andy, she decrypts the cards immediately and gets back the original card numbers.

Becky’s 5-card hand: $g_b(m_j^{b_0}) \equiv (m_j^{b_0})^{b_1} \equiv m_j \ (\text{mod} \ p)$

Now each of the players has a 5-card hand that is only known to himself or herself. If they need to select new cards from the deck, they can follow the same back-and-forth procedures of encrypting and decrypting.

How fair is the poker game played in this manner? How secure is the game? It is very fair and secure if the players follow the rules and do not cheat. It is obviously possible to cheat. When Andy passes the 52 encrypted card numbers to Becky, Becky certainly can try to break Andy’s lock by figuring out Andy’s $a_0$. When Becky passes her encrypted cards to Andy, he can try to figure out Becky’s $b_0$. For that to happen, the player who wants to cheat will need to have enormous amount of computational resources at the ready. Thus the prime number $p$ should be large enough to make cheating an intractable problem. On the other hand, even when the prime number is of a moderate size, there has to be a fair amount of computational resources in order to play the game efficiently, with all the encrypting and decrypting that have to be done.

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Fermat’s Little Theorem

We now use Fermat’s Little Theorem to show that the encryption-decryption key works correctly and accurately. We show the following:

$(m^{a_0})^{a_1} \equiv m \ (\text{mod} \ p) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

For the descriptions of the numbers $m$, $p$, $a_0$ and $a_1$, see the above section Setting Up the Padlocks. First we state the Fermat’s Little Theorem.

Fermat’s Little Theorem
Let $q$ be a prime number. Then for any integer $a$, $a^q-a$ is an integer multiple of $q$ (or $q$ divides $a^q-a$). Using congruence notation, the theorem can be expressed as:

$a^q \equiv a \ (\text{mod} \ q)$

If the integer $a$ is not divisible by $q$, then we can divide out $a$ and the theorem can be expressed as:

$a^{q-1} \equiv 1 \ (\text{mod} \ q)$

For a proof and a fuller discussion of Fermat’s little theorem, see this post.

We now prove the property $(1)$. Recall that the pair of positive integers $a_0$ and $a_1$ are keys to lock and unlock a number $m$. They are chosen such that $a_0 \cdot a_1 \equiv 1 \ (\text{mod} \ p-1)$, or equivalently $a_0 \cdot a_1=1+(p-1) \cdot k$ for some integer $k$. This integer $k$ must be positive since $a_0$ and $a_1$ are both positive.

In the derivation below, we repeated use the fact that $m^p \equiv m \ (\text{mod} \ p)$ (applying the Fermat’s Little Theorem).

\displaystyle \begin{aligned} m&\equiv m^p \ (\text{mod} \ p)=m \cdot m^{p-1} \\&\equiv m^p \cdot m^{p-1} \ (\text{mod} \ p)=m \cdot m^{2(p-1)} \\&\equiv m^p \cdot m^{2(p-1)} \ (\text{mod} \ p)=m \cdot m^{3(p-1)} \\&\cdots \\&\cdots \\&\cdots \\&\equiv m^p \cdot m^{(k-1)(p-1)} \ (\text{mod} \ p)=m \cdot m^{k(p-1)} \\&=m \cdot m^{k(p-1)} \equiv m^{a_0 \cdot a_1} \ (\text{mod} \ p) \end{aligned}

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Example of Congruence Calculation

For a numerical example, we use a small prime number $p=55,049$. Though a small prime number, it is large enough to make the illustration meaningful. Andy chooses $a_0 \cdot a_1$ such that $a_0 \cdot a_1=1+(p-1) \cdot k$ for some integer $k$. Andy decides to use $k=3817$, leading to $a_0=2,657$ and $a_1=79,081$.

As illustration of how the calculation is done, let $m=1020$ (the number for $\heartsuit 2$ as indicated above).

To decrypt this card, Andy needs to raise $1020$ to the 2657th power and then find the remainder upon division by $p=50,049$. This is the definition for $1010200^{269}$ modulo $p$. But the calculation is not easy to do directly without special software. We present here a “divide and conquer” approach that use the division algorithm in each step to reduce the exponent by half.

To start, note that $1020^2 \equiv 49518 \ (\text{mod} \ 55049)$, meaning that the remainder is $49518$ when $1020^2$ is divided by $55049$. In the following series of steps, a congruence calculation is performed in each step (using the division algorithm) to reduce the exponent by half.

\displaystyle \begin{aligned} 1020^{2657}&\equiv (1020^2)^{1328} \cdot 1020 \ (\text{mod} \ 55049) \\&\text{ } \\&\equiv 49518^{1328} \cdot 1020 \equiv (49518^2)^{664} \cdot 1020 \\&\text{ } \\& \equiv 39766^{664} \cdot 1020 \equiv (39766^2)^{332} \cdot 1020 \\&\text{ } \\& \equiv 52231^{332} \cdot 1020 \equiv (52231^2)^{166} \cdot 1020 \\&\text{ } \\&\equiv 14068^{166} \cdot 1020 \equiv (14068^2)^{83} \cdot 1020 \\&\text{ } \\& \equiv 7469^{83} \cdot 1020 \equiv (7469^2)^{41} \cdot 7469 \cdot 1020 \\&\text{ } \\& \equiv 21324^{41} \cdot 7469 \cdot 1020 \\&\text{ } \\&\equiv 21324^{41} \cdot 21618 \equiv (21324^2)^{20} \cdot 21324 \cdot 21618 \\&\text{ } \\& \equiv 8236^{20} \cdot 21324 \cdot 21618 \\&\text{ } \\& \equiv 8236^{20} \cdot 1906 \equiv (8236^2)^{10} \cdot 1906 \\&\text{ } \\& \equiv 11328^{10} \cdot 1906 \equiv (11328^2)^{5} \cdot 1906 \\&\text{ } \\& \equiv 4365^5 \cdot 1906 \equiv (4365^2)^2 \cdot 4365 \cdot 1906 \\&\text{ } \\& \equiv 6271^2 \cdot 4365 \cdot 1906 \\&\text{ } \\&\equiv 6271^2 \cdot 7291 \\&\text{ } \\&\equiv 20455 \cdot 7291 \\&\text{ } \\&\equiv 9664 \ (\text{mod} \ 55049) \end{aligned}

Thus the card number $1020$ is encrypted as $9664$. To recover the original card number from this encrypted number, Andy needs to raise $9664$ to the power of $a_1=79081$. Here, we get an assist from Fermat’s Little Theorem in addition to the ‘divide and conquer” congruence arithmetic that is used above.

According to Fermat’s Little Theorem, $9664^{55048} \equiv 1 \ (\text{mod} \ 55049)$. Thus we have

$9664^{79081} \equiv 9664^{55048} \cdot 9664^{24033} \equiv 9664^{24033} \ (\text{mod} \ 55049)$

With the help of Fermat’s Little Theorem, the exponent $79081$ has come down to $24033$. In the rest of the way, the “divide and conquer” approach is used.

\displaystyle \begin{aligned} 9664^{24033}&\equiv (9664^2)^{12016} \cdot 9664 \ (\text{mod} \ 55049) \\&\text{ } \\&\equiv 29782^{12016} \cdot 9664 \equiv (29782^2)^{6008} \cdot 9664 \\&\text{ } \\&\equiv 8237^{6008} \cdot 9664 \equiv (8237^2)^{3004} \cdot 9664 \\&\text{ } \\&\equiv 27801^{3004} \cdot 9664 \equiv (27801^2)^{1502} \cdot 9664 \\&\text{ } \\&\equiv 7641^{1502} \cdot 9664 \equiv (7641^2)^{751} \cdot 9664 \\&\text{ } \\&\equiv 32941^{751} \cdot 9664 \equiv (32941^2)^{375} \cdot 32941 \cdot 9664 \\&\text{ } \\&\equiv 38642^{375} \cdot 32941 \cdot 9664 \\&\text{ } \\&\equiv 38642^{375} \cdot 48506 \equiv (38642^2)^{187} \cdot 38642 \cdot 48506 \\&\text{ } \\&\equiv 39^{187} \cdot 38642 \cdot 48506 \\&\text{ } \\&\equiv 39^{187} \cdot 5451 \equiv (39^2)^{93} \cdot 39 \cdot 5451 \\&\text{ } \\&\equiv 1521^{93} \cdot 39 \cdot 5451 \\&\text{ } \\&\equiv 1521^{93} \cdot 47442 \equiv (1521^2)^{46} \cdot 1521 \cdot 47442 \\&\text{ } \\&\equiv 1383^{46} \cdot 1521 \cdot 47442 \\&\text{ } \\&\equiv 1383^{46} \cdot 45092 \equiv (1383^2)^{23} \cdot 45092 \\&\text{ } \\&\equiv 41023^{23} \cdot 45092 \equiv (41023^2)^{11} \cdot 41023 \cdot 45092 \\&\text{ } \\&\equiv 38599^{11} \cdot 41023 \cdot 45092 \\&\text{ } \\&\equiv 38599^{11} \cdot 52618 \equiv (38599^2)^{5} \cdot 38599 \cdot 52618 \\&\text{ } \\&\equiv 36665^5 \cdot 38599 \cdot 52618 \\&\text{ } \\&\equiv 36665^5 \cdot 24376 \equiv (36665^2)^{2} \cdot 36665 \cdot 24376 \\&\text{ } \\&\equiv 25645^2 \cdot 36665 \cdot 24376 \\&\text{ } \\&\equiv 25645^2 \cdot 25525 \\&\text{ } \\&\equiv 50671 \cdot 25525 \\&\text{ } \\&\equiv 1020 \ (\text{mod} \ 55049) \end{aligned}

In each step of the above calculation, the division algorithm is applied to reduce the exponent by half. For example, to go from the first line to the second line, $9664^2$ is divided by $55049$ to obtain the remainder $29782$, i.e. $9664^2 \equiv 29782 \ (\text{mod} \ 55049)$. The number $1020$ in the last line is the remainder when $50671 \cdot 25525$ is divided by $55049$.

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$\copyright \ 2013 \text{ by Dan Ma}$

# Fermat’s Little Theorem and RSA Algorithm

RSA is a cryptographic algorithm that is used to send and receive messages. We use the Fermat’s Little Theorem to prove that RSA works correctly and accurately. In other words, the decrypted message is indeed the original message from the sender. Mathematically we show that applying the encryption function and the decryption function successively produces the identity function.

To see how RSA works, see the previous post An Illustration of the RSA Algorithm.

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RSA Algorithm

We first briefly describe the algorithm and then present the mathematical statement to validate.

Let $N=p \cdot q$ where $p$ and $q$ are two prime numbers. Let $\phi=(p-1) \cdot (q-1)$. Choose an integer $e$ with $1 such that $e$ and $\phi$ are relatively prime.

The public key consists of $N$ and $e$ where $e$ is the encryption key. Once it is published, anyone can use it to encrypt messages to send to the creator of the public key. The following is the encryption function:

$f(M) \equiv M^e \ (\text{mod} \ N)$

where $M$ is a positive integer and is the original message.

The private key is a positive integer $d$ that satisfies:

$d \cdot e \equiv 1 \ (\text{mod} \ \phi=(p-1) \cdot (q-1))$

In other words, $d$ is the multiplicative inverse of $e$ in the modular arithmetic of modulo $\phi$. The above condition is equivalent to: $de-1=(p-1) \cdot (q-1) \cdot k$ for some integer $k$.

The number $d$ is the decryption key that will be used to decode messages. So it should remain private.

Once the creator of the public key receives an encrypted message $C=f(M)$, he or she uses the following decryption function to obtain the original message $M$.

$g(C) \equiv C^d \ (\text{mod} \ N)$

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The Mathematical Statement to Validate

What we prove is that the decryption function is to undo the encryption function. Specifically, we prove the following:

$g(C)=g(f(M))=(M^e)^d=M^{ed} \equiv M \ (\text{Mod} \ N) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

In other words, applying the decryption function $g$ to the encryption function $f$ produces the original message.

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Fermat’s Little Theorem

In this section, we list out the tools we need to prove the correctness of RSA.

Theorem 1 (Fermat’s Little Theorem)
If $p$ is a prime number and $a$ is an integer such that $a$ and $p$ are relatively prime, then

$a^{p-1}-1$ is an integer multiple of $p$

or equivalently $a^{p-1} \equiv 1 \ (\text{mod} \ p)$.

For a proof of Fermat’s little theorem, see this post.

Lemma 2 (Euclid’s Lemma)
Let $a$, $b$ and $d$ be integers where $d \ne 0$. Then if $d$ divides $a \cdot b$ (symbolically $d \lvert a \cdot b$), then either $d \lvert a$ or $d \lvert b$.

Euclid’s Lemma is needed to prove the following Lemma.

Lemma 3
Let $M$ be an integer. Let $p$ and $q$ be prime numbers with $p \ne q$.

Then if $a \equiv M \ (\text{mod} \ p)$ and $a \equiv M \ (\text{mod} \ q)$, then $a \equiv M \ (\text{mod} \ p \cdot q)$.

Proof of Lemma 3
Suppose we have $a \equiv M \ (\text{mod} \ p)$ and $a \equiv M \ (\text{mod} \ q)$. Then for some integers $i$ and $j$, we have:

$a=M+p \cdot i$ and $a=M+q \cdot j$.

Then $p \cdot i=q \cdot j$. This implies that $p$ divides $q \cdot j$ ($p \lvert q \cdot j$). By Euclid’s lemma, we have either $p \lvert q$ or $p \lvert j$. Since $p$ and $q$ are distinct prime numbers, we cannot have $p \lvert q$. So we have $p \lvert j$ and that $j=p \cdot w$ for some integer $w$.

Now, $a=M+q \cdot j=M+q \cdot p \cdot w$, implying that $a \equiv M \ (\text{mod} \ p \cdot q)$. $\blacksquare$

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The Proof of (1)

We now prove the property $(1)$ described above. We show that

$(M^e)^d=M^{ed} \equiv M \ (\text{Mod} \ N=p \cdot q)$

We first show that $M^{ed} \equiv M \ (\text{Mod} \ p)$ and $M^{ed} \equiv M \ (\text{Mod} \ q)$. Then the desired result follows from Lemma 3.

To show $M^{ed} \equiv M \ (\text{Mod} \ p)$, we consider two cases: $M \equiv 0 \ (\text{Mod} \ p)$ or $M \not \equiv 0 \ (\text{Mod} \ p)$.

Case 1. $M \equiv 0 \ (\text{Mod} \ p)$. Then $M$ is an integer multiple of $p$, say $M=p \cdot w$ where $w$ is an integer. Then $M^{ed}=(p \cdot w)^{ed}=p \cdot p^{ed-1} \cdot w^{ed}$. So both $M$ and $M^{ed}$ are integer multiples of $p$. Thus $M^{ed} \equiv M \ (\text{Mod} \ p)$.

Case 2. $M \not \equiv 0 \ (\text{Mod} \ p)$. This means that $p$ and $M$ are relatively prime (having no common divisor other than 1). Thus we can use Fermat’s Little Theorem. We have $M^{p-1} \equiv 1 \ (\text{mod} \ p)$.

From the way the decryption key $d$ is defined above, we have $ed-1=(p-1) \cdot (q-1) \cdot k$ for some integer $k$. We then have:

\displaystyle \begin{aligned} M^{ed}&=M^{ed-1} \cdot M \\&=M^{(p-1) \cdot (q-1) \cdot k} \cdot M \\&=(M^{p-1})^{(q-1) \cdot k} \cdot M \\&\equiv (1)^{(q-1) \cdot k} \cdot M \ (\text{Mod} \ p) \ * \\&\equiv M \ (\text{Mod} \ p) \end{aligned}

At the step with *, we apply Fermat’s Little Theorem. So we have $M^{ed} \equiv M \ (\text{Mod} \ p)$.

The same reason reasoning can show that $M^{ed} \equiv M \ (\text{Mod} \ q)$.

By Lemma 3, it follows that $M^{ed} \equiv M \ (\text{Mod} \ N=p \cdot q)$. $\blacksquare$

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$\copyright \ 2013 \text{ by Dan Ma}$

Revised August 9, 2014.