# Fermat’s Little Theorem as a primality testing

Armed with an understanding a theorem of Fermat and a calculator or software that can handle large numbers, any interested individual can carry out exercises in primality testing – checking whether or not a number is prime. No advanced study in number theory is required. In this post, we explain how this is done. This post is built on two previous posts on detecting composite numbers and the fast powering algorithm.

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Fermat’s Little Theorem

This is called the little theorem so that it is not confused with the other theorem of Fermat whose proof took three and a half centuries. The little theorem of Fermat describes a property that is possessed by all prime numbers. It says: if $n$ is a prime number, then a certain property is true for the number $n$. Turning this around, if that property is not true for a number $n$, that number $n$ must not be prime (i.e. it must be composite). This approach is a great way to prove the compositeness of a number. It is also a great way to gather evidence to support the primality of a number.

Two number are relatively prime to each other if they do not have any common factor that is greater than 1. For example, 28 is relatively prime to 15 even though neither one of them is prime. Similarly 17 and 35 have no factors in common. So they are relatively prime to each other.

The notation $P \equiv Q \ (\text{mod} \ n)$ means that $P$ and $Q$ differ by a multiple of $n$, i.e. $P-Q=n k$ for some integer $k$. The notation is read $P$ is congruent to $Q$ modulo $n$. The statement of Fermat’s theorem uses the notation $a^{n-1} \equiv 1 \ (\text{mod} \ n)$. This would mean that $a^{n-1}-1=n k$ for some integer $k$. This means that $a^{n-1}-1$ divisible by $n$.

Here’s the statement of Fermat’s little theorem.

Fermat’s Little Theorem
If $n$ is a prime number, then $a^{n-1} \equiv 1 \ (\text{mod} \ n)$ for all numbers $a$ that are relatively prime to $n$.

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Fermat’s Test

The theorem says that if $n$ is a prime number, $2^{n-1} \equiv 1 \ (\text{mod} \ n)$ and $3^{n-1} \equiv 1 \ (\text{mod} \ n)$ and so on. The flip side is that if we can find a number $a$ such that $a^{n-1} \not \equiv 1 \ (\text{mod} \ n)$, then we have convincing proof that $n$ is not a prime number.

When we need to determine whether or not a number $n$ is prime, we pick a value $a$ and calculate $a^{n-1} \ (\text{mod} \ n)$. If the calculated result is not 1, then we have a convincing proof that the number $n$ is not prime. If $a^{n-1} \equiv 1 \ (\text{mod} \ n)$, then the number $n$ is said to be a probable prime to base $a$. When $n$ is a probable prime to base $a$, we also say that $n$ passes the Fermat-a test. Fermat’s little theorem can then be restated as:

Fermat’s Little Theorem
If $n$ is a prime number, then $n$ passes Fermat-a test for all possible choices of $a$.

Example 1
This previous post discusses the following 204-digit number.

$N=$
34526903293921580314641092817369674040684481568423
96721012992064214519445919256941544565276067662360
10874972724155570842527652727868776362959519620872
73561220060103650687168112461098659687818073890148
6527

It can be verified that $2^{N-1} \equiv W \ (\text{mod} \ N)$ where $W$ is not 1 and is the following 203-digit number.

$W=$
33345811005959530251539697392827903173946066773819
70645616725285996925661000056829272733579262095715
97827398131150054514508640724258354848985651127636
92970799269335402819507605691622173717318335512037
458

The number $N$ in this example fails the Fermat-2 test and is thus a composite number. The calculation of $2^{N-1} \equiv W \ (\text{mod} \ N)$ is carried out by using the fast powering algorithm.

Example 2
Consider $M=$ 982447387. Is it prime?

Calculate $a^{M-1} \ (\text{mod} \ M)$ for several values of $a$. let’s start with 2. We find that $2^{M-1} \equiv 1 \ (\text{mod} \ M)$. So we say that the number $M$ is a probable prime to base 2. It is very likely that $M$ is a prime. Let’s try a few more values of $a$. We have the following results.

$3^{M-1} \equiv 1 \ (\text{mod} \ M)$

$5^{M-1} \equiv 1 \ (\text{mod} \ M)$

$7^{M-1} \equiv 1 \ (\text{mod} \ M)$

Now $M$ is a probable prime to the bases 2, 3, 5, and 7. It turns out that there are actually composite numbers that are probable primes to bases 2, 3, 5 and 7. But they are extremely rare. So the number $M$ is a probable prime (meaning that the chance that it is not a prime is very negligible). For good measure, we generate 5 random values of $a$. They are:

323064262
866079494
599071122
773037193
247321577

Calculating $a^{M-1} \ (\text{mod} \ M)$ for all 5 values show that the number $M$ passes the Fermat-a test for all 5 values of $a$. At this point we are willing to conclude that $M$ is a prime. The conclusion is based on a probability argument and is not based on a mathematical proof.

Since the number $M$ is a 9-digit number, we can actually conclude that it is a prime by looking it up in lists of primes that are published online. So this is a toy example that is meant for illustration only.

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Remarks

Example 1 shows that failing the Fermat-a test just for one value of $a$ is enough to conclude that the number in question is not prime. In general, to test the primality of a number $n$, carry out the Fermat-a test for several values of $a$ (preferably randomly chosen). If $n$ fails the Fermat-a test for one of the values of $a$, then $n$ is composite. On the other hand, if $n$ passes the Fermat-a test for all the chosen values of $a$, then $n$ is a probable prime.

The Fermat test is the “entry level” primality test. It is easy to use. It often works correctly. But it can produce incorrect results on rare occasions. For a more in depth discussion of the Fermat’s primality test, we refer you to an article in another blog.

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$\copyright \ 2015 \text{ by Dan Ma}$

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# An easy method for detecting composite numbers

How do you test whether a number is a prime number? Trial division is in use since antiquity and is easy to understand. This is the process of dividing the number in question by each prime number that is smaller than it. If a smaller factor is found, then it is not a prime number (i.e. is a composite number). If no smaller factors can be found, then the number is a prime numbers. However numbers routinely used in encryption algorithms today have hundreds or even thousands of decimal digits. Factoring such large numbers is really a hard problem. If we just want to know whether a number is prime or not, the answer can be obtained very quickly. In this post we showcase a calculation that can be used for checking whether a whole number is prime or composite without actually finding any factor. This method is based on a theorem of Fermat and can be implemented using the fast powering algorithm.

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Examples

Example 1
The author of a recent article came across the following 204-digit number that he incorrectly asserted as a prime number.

$N=$
34526903293921580314641092817369674040684481568423
96721012992064214519445919256941544565276067662360
10874972724155570842527652727868776362959519620872
73561220060103650687168112461098659687818073890148
6527

It turns out that showing the number $N$ is composite is not difficult. The idea is what can be called the Fermat-2 test: if $N$ is a prime number, then $2^{N-1} \equiv 1 \ (\text{mod} \ N)$, meaning that $2^{N-1}-1$ is divisible by $N$, or equivalently the remainder is 1 when $2^{N-1}$ is divided by $N$. Using the fast powering algorithm, the exponentiation $2^{N-1}$ modulo $N$ is turned into a series of squarings and multiplying. The result is that $2^{N-1} \equiv W \ (\text{mod} \ N)$ where $W$ is the following 203-digit number:

$W=$
33345811005959530251539697392827903173946066773819
70645616725285996925661000056829272733579262095715
97827398131150054514508640724258354848985651127636
92970799269335402819507605691622173717318335512037
458

If $N$ were a prime number, then $2^{N-1}$ modulo $N$ is 1. But $2^{N-1}$ modulo $N$ is clearly not 1. Therefore $N$ must be composite. We do not show the steps that produce the 203-digit number $W$. In carrying out the square-and-multiply algorithm, there are 676 squarings and 327 multiplications. The process runs quickly when it is implemented on a computer. In the next example, we work a small example to illustrate how the fast powering algorithm works.

Example 2
Let’s do a smaller example to demonstrate the idea of the Fermat-2 test and the calculation of the fast powering algorithm (also called the square-and-multiply algorithm). Consider the number 55289. Is it a prime number?

We show that $2^{55288} \equiv 8349 \ (\text{mod} \ 55289)$. So 55289 is not a prime (otherwise $2^{55288}$ would be 1 modulo 55289). To show $2^{55288} \ (\text{mod} \ 55289)$, the first step is to express the exponent 55288 in its binary expansion.

$55288=2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}+2^{12}+2^{14}+2^{15}$

Step 2 is to perform a series of 15 squarings (15 is the highest power of 2 in the binary expansion of the exponent 55288). Step 3 is to perform a series of multiplications. Both steps are shown in the following table.

Example 2 Results

$\left[\begin{array}{rrrrr} i & \text{ } & \text{Squaring} & \text{ } & \text{Multiplying} \\ \text{ } & \text{ } \\ 0 & \text{ } & 2 & \text{ } & \text{ } \\ 1 & \text{ } & 4 & \text{ } & \text{ } \\ 2 & \text{ } & 16 & \text{ } & \text{ } \\ 3 & \text{ } & 256* & \text{ } & 256 \\ 4 & \text{ } & 10247* & \text{ } & 24649 \\ 5 & \text{ } & 7198* & \text{ } & 1101 \\ 6 & \text{ } & 5411* & \text{ } & 41588 \\ 7 & \text{ } & 31040* & \text{ } & 3948 \\ 8 & \text{ } & 15486* & \text{ } & 44383 \\ 9 & \text{ } & 27803* & \text{ } & 40647 \\ 10 & \text{ } & 11300* & \text{ } & 25377 \\ 11 & \text{ } & 27699 & \text{ } & \text{ } \\ 12 & \text{ } & 44437* & \text{ } & 3305 \\ 13 & \text{ } & 334 & \text{ } & \text{ } \\ 14 & \text{ } & 978* & \text{ } & 25528 \\ 15 & \text{ } & 16571* & \text{ } & 8349 \end{array}\right]$

The column for squaring starts with 2, the base of the exponentiation $2^{55288}$. Each number in that column is the square of the preceding number and is reduced modulo 55289. The numbers with asterisks refer to the positions that are 1s in the binary expansion of 55288. The third column shows the multiplications of the numbers with asterisks in the second column.

Example 3
Consider the following 309-digit number:

RSA-1024
13506641086599522334960321627880596993888147560566
70275244851438515265106048595338339402871505719094
41798207282164471551373680419703964191743046496589
27425623934102086438320211037295872576235850964311
05640735015081875106765946292055636855294752135008
52879416377328533906109750544334999811150056977236
890927563

This example is an RSA number called RSA-1024. It is a 1024-bit (309-digit in decimal). It is a product of two prime numbers and thus is not a prime number. Indeed, $2^{N-1} \equiv T \ (\text{mod} \ N)$ where $N$ is RSA-1024 and $T$ is the following number:

$T=$
12093909443203361586765059535295699686754009846358
89512389028083675567339322020593385334853414711666
28419681241072885123739040710771394053528488357104
98409193003137847878952260296151232848795137981274
06300472693925500331497519103479951096634123177725
21248297950196643140069546889855131459759160570963
857373851

Clearly, RSA-1024 does not pass the Fermat-2 test and is thus a composite number. Yet RSA-1024 has not yet been factored. It is not expected to be factored in decades to come barring a dramatic breakthrough in computing technology. This example points to the principle on which the RSA cryptosystem is based, that

It is relatively easy to decide whether or not a number is prime. But it is hard to find the prime factors of a given composite number.

In short, primality testing is easy while factoring is hard. In any case, the examples of RSA-1024 and other RSA numbers are empirical evidence that factoring is a hard problem.

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Remarks

The Fermat-2 test used here is based on Fermat’s Little Theorem, which is discussed in the next post.

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$\copyright \ 2015 \text{ by Dan Ma}$

# A fast way to do modular exponentiation

In number theory and in cryptography, it is often necessary to raise a number $a$ to a number $e$ modulo another number $N$ (the number $e$ here stands for exponent and is not to be confused with the natural log constant). In this post we explain an algorithm that can do such exponentiation efficiently. This algorithm goes by many names. Some of the common ones are fast powering algorithm, fast modular exponentiation, and square and multiply.

The next post shows how the fast powering algorithm is used in the context of primality testing (i.e. checking whether or not a number is prime).

The usual notation for raising a number $a$ to a number $e$ modulo another number $N$ is $a^{e} \ (\text{mod} \ N)$. The answer to this exponentiation is in effect the remainder obtained by dividing the number $a^e$ by $N$. For example the answer of $2^{6} \ (\text{mod} \ 11)$ is 9. When $2^6=$ 64 is divided by 11, the remainder is 9. We use the notation $2^{6} \equiv 9 \ (\text{mod} \ 11)$ to express this result. In cryptography, the exponent $e$ and the modulo $N$ are large numbers with hundreds or even thousands of decimal digits. For example, it can be verified that $c^{d} \equiv m \ (\text{mod} \ N)$ where

$N=$
11438162575788886766923577997614661201021829672124
23625625618429357069352457338978305971235639587050
58989075147599290026879543541

$c=$
96869613754622061477140922254355882905759991124574
31987469512093081629822514570835693147662288398962
8013391990551829945157815154

$d=$
10669861436857802444286877132892015478070990663393
78628012262244966310631259117744708733401685974623
06553968544513277109053606095

$m=$
20080500130107090300231518041900011805001917210501
1309190800151919090618010705

The modulus $N$ is a 129-digit number that is used as the modulus of a factoring challenge problem called RSA-129 challenge. The challenge was to factor the number $N$ into its prime factors. Knowing the prime factors of the modulus $N$ means that any message encrypted using that modulus will no longer be secure. The 129-digit $N$ was factored in 1994. So the RSA encryption scheme using any modulus similarly in size is not secure. So the calculation as shown above is actually considered a toy problem, though a more realistic one. The fast powering algorithm (or square-and-multiply algorithm) discussed here can be used to calculate $c^{d} \equiv m \ (\text{mod} \ N)$.

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Examples

Example 1
To illustrate how fast powering works, we start with a small example. We perform $3^{23} (\text{mod} \ 29)$. In the fast powering approach, we do not need to raise 3 to 23. Instead, we first find the binary expansion of the exponent 23 to transform the computation of $3^{23}$ into a series of squarings and multiplications. To this end, we write $23$ as a sum of powers of two as follows:

$23=2^0+2^1+2^2+2^{4}$

Next we compute $3^{2^0},3^{2^1},3^{2^2},3^{2^3},3^{2^{4}}$ modulo $29$. Note that each term is the square of the preceding term, hence the word square in the name “square-and-multiply”. The following shows the squarings, all modulo 29:

\displaystyle \begin{aligned} &3^{2^0} \equiv 3 \ * \\&3^{2^1} \equiv 9 \ * \\&3^{2^2} \equiv 9^2 \equiv 23 \ * \\&3^{2^3} \equiv 23^2 \equiv 7 \\&2^{2^{4}} \equiv 7^2 \equiv 20 \ * \end{aligned}

Note that the squarings marked by * are the powers of 2 in the binary expansion of 23. In the above series of squarings, there are 4 multiplications and 4 divisions for the reduction modulo $29$. The next step is to multiply the numbers with *. After each multiplication, reduce modulo 29.

\displaystyle \begin{aligned} 3^{23} &\equiv 3 \cdot 9 \cdot 23 \cdot 20 \\&\equiv 27 \cdot 23 \cdot 20 \\&\equiv 12 \cdot 20 \\&\equiv 8 \ \text{mod} \ 29 \end{aligned}

Thus the answer is $3^{23} \equiv 8 \ (\text{mod} \ 29)$. To simplify the presentation of the calculation, we use a table such as the one below. The middle column shows the squarings. The asterisks in the squaring column indicate that the results come from the powers of 2 in the binary expansion of the exponent, meaning that these will be the numbers to be multiplied in the third column.

Example 1 Results

$\left[\begin{array}{lllll} i & \text{ } & \text{Squaring} & \text{ } & \text{Multiplying} \\ \text{ } & \text{ } \\ 0 & \text{ } & 3* & \text{ } & 3 \\ 1 & \text{ } & 9* & \text{ } & 27 \\ 2 & \text{ } & 23* & \text{ } & 12 \\ 3 & \text{ } & 7 & \text{ } & \text{ } \\ 4 & \text{ } & 20* & \text{ } & 8 \end{array}\right]$

Example 2
Verify the following

$30120^{17} \equiv 23877 \ (\text{mod} \ 44197)$

$23877^{41201} \equiv 30120 \ (\text{mod} \ 44197)$

This is a toy example of RSA encryption (the first calculation) and decryption (the second). The modulus $N=$ 44197 and the first exponent 17 are the public key. The first exponentiation turns the message 30120 into the ciphertext 23877. The second exponentiation converts the ciphertext back into the original message. We demonstrate how to do the decryption (the second calculation). First express the exponent 41201 into its binary expansion.

$41201=2^0+2^4+2^5+2^{6}+2^{7}+2^{13}+2^{15}$

The following table shows the squarings and multiplications.

Example 2 Results

$\left[\begin{array}{lllll} i & \text{ } & \text{Squaring} & \text{ } & \text{Multiplying} \\ \text{ } & \text{ } \\ 0 & \text{ } & 23877* & \text{ } & 23877 \\ 1 & \text{ } & 14026 & \text{ } & \text{ } \\ 2 & \text{ } & 7829 & \text{ } & \text{ } \\ 3 & \text{ } & 36199 & \text{ } & \text{ } \\ 4 & \text{ } & 14945* & \text{ } & 39384 \\ 5 & \text{ } & 25584* & \text{ } & 41247 \\ 6 & \text{ } & 27683* & \text{ } & 11206 \\ 7 & \text{ } & 16706* & \text{ } & 33141 \\ 8 & \text{ } & 30578 & \text{ } & \text{ } \\ 9 & \text{ } & 26549 & \text{ } & \text{ } \\ 10 & \text{ } & 39842 & \text{ } & \text{ } \\ 11 & \text{ } & 5512 & \text{ } & \text{ } \\ 12 & \text{ } & 18805 & \text{ } & \text{ } \\ 13 & \text{ } & 7828* & \text{ } & 35555 \\ 14 & \text{ } & 20542 & \text{ } & \text{ } \\ 15 & \text{ } & 25005* & \text{ } & 30120 \end{array}\right]$

The modulus in this example is small enough so that all the reduction modulo $N$ can be done by a hand-held calculator. For example, take the first squaring. The square $23877^2$ divided by 44197 leads to the quotient 12899. Then $23877^2-12899 \cdot 44197$ produces the remainder 14026. For a larger modulus $N$, use a calculator or software that can handle multiplication and divisions of larger numbers.

Example 3
This is a slightly larger example that is similar to Example 2. Verify the following

$230914^{17} \equiv 347965 \ (\text{mod} \ 497347)$

$347965^{204209} \equiv 230914 \ (\text{mod} \ 497347)$

We perform the second exponentiation. The following is the binary expansion of the exponent 204209.

$204209=2^0+2^4+2^5+2^{7}+2^{8}+2^{10}+2^{11}+2^{12}+2^{16}+2^{17}$

The following table shows the squarings and multiplications.

Example 3 Results

$\left[\begin{array}{lllll} i & \text{ } & \text{Squaring} & \text{ } & \text{Multiplying} \\ \text{ } & \text{ } \\ 0 & \text{ } & 347965* & \text{ } & 347965 \\ 1 & \text{ } & 16728 & \text{ } & \text{ } \\ 2 & \text{ } & 316970 & \text{ } & \text{ } \\ 3 & \text{ } & 416083 & \text{ } & \text{ } \\ 4 & \text{ } & 64230* & \text{ } & 12464 \\ 5 & \text{ } & 496882* & \text{ } & 172404 \\ 6 & \text{ } & 216225 & \text{ } & \text{ } \\ 7 & \text{ } & 145890* & \text{ } & 187076 \\ 8 & \text{ } & 424582* & \text{ } & 299597 \\ 9 & \text{ } & 486410 & \text{ } & \text{ } \\ 10 & \text{ } & 254689* & \text{ } & 88899 \\ 11 & \text{ } & 4246* & \text{ } & 476128 \\ 12 & \text{ } & 124024* & \text{ } & 295068 \\ 13 & \text{ } & 4560 & \text{ } & \text{ } \\ 14 & \text{ } & 402373 & \text{ } & \text{ } \\ 15 & \text{ } & 175484 & \text{ } & \text{ } \\ 16 & \text{ } & 400057* & \text{ } & 200467 \\ 17 & \text{ } & 333343* & \text{ } & 230914 \\ \end{array}\right]$

Example 4
This is the RSA-129 challenge problem mentioned earlier. We do not perform the calculation here. The discussion in the running time section below shows that this example requires at most 258 multiplications even though this example is so much larger than the other three examples. The RSA-129 challenge problem is also discussed in another blog authored by the author of this blog.

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Steps in the algorithm

To summarize, there are three steps in carry out the fast exponentiation for $a^{e} \ (\text{mod} \ N)$. They are:

1. Find the binary expansion of the exponent $e$.
2. Perform the series of squarings $a^{2^0}, a^{2^1}, a^{2^2}, \cdots$ up to the highest power of 2 in the binary expansion of the exponent. The result of each squaring is immediately reduced modulo $N$.
3. Multiply the results of the squarings that correspond to the powers of 2 in the binary expansion of the exponent $e$. The result of each multiplication is immediately reduced modulo $N$. The last multiplication result is the answer to the modular exponentiation.

Note that the squarings in the second step are shown in the second column in the above tables. The multiplications are performed in the third column but only for the rows that are the powers of 2 in the binary expansion of the exponent (the rows with *).

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Running time

Let’s look at the running time of the fast powering algorithm. Suppose that $k$ is the highest power of 2 in the exponent $E$ in the exponentiation $a^{E} \ (\text{mod} \ N)$. So we have:

$\displaystyle 2^k \le E \le 2^{k+1}-1$

The above inequality means that the exponent $E$ is a $k+1$-bit number. The above three examples demonstrate that the squaring step requires exactly $k$ multiplications and that the multiply step requires at most $k$ multiplications. Overall, the algorithm requires at most $2k$ many multiplications and the same number of divisions to reduce modulo $N$. Since $2^k \le E$, by taking natural log of both sides we have:

$\displaystyle k \le \frac{\text{log}(E)}{\text{log}(2)}$

$\displaystyle 2k \le 2 \ \frac{\text{log}(E)}{\text{log}(2)}$

This means that the algorithm takes at most $\displaystyle 2 \ \frac{\text{log}(E)}{\text{log}(2)}$ multiplications. In particular, if $E$ is roughly $2^n$ in size, then the fast powering algorithm requires at most $\displaystyle 2 \ \frac{\text{log}(2^{n})}{\text{log}(2)}=2n$ multiplications. For example, the exponent in Example 4 is roughly $2^{129}$ in size. Thus the fast powering algorithm will take at most 258 multiplications for that example. Another example is when the exponent $E$ is a 1024-bit number (309 decimal digits) and is thus roughly $2^{1024}$. Then the algorithm requires at most 2048 multiplications. Thus the fast powering algorithm is fast and efficient.

The next post shows how the fast powering algorithm is used in the context of primality testing (i.e. checking whether or not a number is prime).

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$\copyright \ 2015 \text{ by Dan Ma}$