Recall Fermat’s Theorem from the lecture 09

If \(p\) is prime, \(n \in \mathbb{Z}\), then $$ \begin{align*} a^p &\equiv a \bmod p \\ a^{p-1} &\equiv 1 \bmod p \quad \text{when } (a,p)=1 \end{align*} $$

There is another important definition that we also need as follows

The order of \(a \bmod p\) is the smallest \(n > 0\) with $$ \begin{align*} a^n \equiv 1 \bmod p \end{align*} $$

For example, take \(p = 7\), then

$$ \begin{align*} 1 &\equiv 1 \bmod 7 \quad o(a) = 1 \\ 2^3 &\equiv 1 \bmod 7 \quad o(a) = 3 \\ 3^6 &\equiv 1 \bmod 7 \quad o(a) = 6 \end{align*} $$

What about negative powers of \(a\)? What is \(a^{-n}\)? well this is just \((a^{-1})^n\). So when can we find the inverse of \(a\)? We know that the inverse of \(a\) will exist only if \((a,p)=1\). This is because we’re trying to solve

$$ \begin{align*} ax \equiv 1 \bmod p \end{align*} $$

This is just \(ax = 1 + py\) and this has a solution if \((a,p) = 1\) by Euclid’s algorithm.

Properties of the order of \(a \bmod p\)

Suppose now that \(p\) is prime and \(n > 0\). We have two very important results:

If \(p\) is prime and \(a > 0\), the order of \(a \bmod p\) must divide any \(n\) such that \(a^n \equiv 1\bmod p\).

Proof: Let’s look at all the possible integers \(n\) that satisfies this. So consider the set

$$ \begin{align*} S = \{n \in \mathbb{Z} \mid a^n \equiv 1 \bmod p \} \end{align*} $$

\(S\) is closed under the addition and subtraction of the exponents since if \(a^m \equiv 1 \bmod p\) and \(a^n \equiv 1 \bmod p\),

$$ \begin{align*} a^ma^n &= a^{m+n} \equiv 1 \bmod p \\ a^ma^{-n} &= a^{m-n} \equiv 1 \bmod p \end{align*} $$

Therefore, \(S\) is an ideal. But we know that any ideal in \(\mathbb{Z}\) is a principle ideal so there has to be some smallest integer such that all of these integers are multiples of it. In other words,

$$ \begin{align*} S = (d) = \{kd \mid k \in \mathbb{Z}\} \end{align*} $$

Therefore, \(d\) is the order of \(a \bmod p\) and for any \(n \in S\), we have \(n = kd\) for some \(k \in \mathbb{Z}\). Thus, \(d \mid n\) such that \(a^n \equiv 1 \bmod p\). \(\ \blacksquare\)

The second result is

If \(p\) is prime and \(a > 0\), the order of \(a \bmod p\) must divide any \(p-1\).

Proof: So we already established that \(d\) divides any \(n\) such that \(a^n \equiv 1 \bmod p\). Now observe that by Fermat’s theorem

$$ \begin{align*} a^{p-1} \equiv 1 \bmod p \end{align*} $$

So this means that \(d\) or the order of \(a\) divides \(p-1\).

Application

One application of this is the following theorem

Suppose \(p\) and \(q\) are primes and that \(p\) divides \(2^{q} - 1\). Then $$ \begin{align*} p &\equiv 1 \bmod q \end{align*} $$

Proof

So we know that \(p\) divides \(2^q - 1\). This means that

$$ \begin{align*} 2^q - 1 \equiv 0 \bmod p \\ 2^q \equiv 1 \bmod p \end{align*} $$

Therefore, the order of \(2\) must divide \(q\) by the previous result. But \(q\) is prime so the order of \(2\) can either be 1 or \(q\). However, it can’t be 1 since \(2 \not\equiv 1 \bmod p\). So the order of 2 must be \(q\). But we also showed using Fermat’s theorem that the order must also divide \(p-1\). Therefore, \(q\) divides \(p - 1\). But this just means

$$ \begin{align*} p - 1 \equiv 0 \bmod q \\ p \equiv 1 \bmod p \end{align*} $$

as we wanted to show. \(\ \blacksquare\)

Example

Show that \(2^{13} - 1\) is prime.

One way to do this, is to iterate through all primes less than \(\sqrt{2^{13}} = \sqrt{8191} = 91\) and see if they divide \(2^{13}\). This works but there is a faster way. We can use the previous theorem we just proved where if \(p\) and \(q\) are primes and \(p\) divides \(2^q - 1\), then \(p \equiv 1 \bmod q\). So if \(p \mid 2^{13} - 1\), then this means

$$ \begin{align*} p \equiv 1 \bmod 13 \end{align*} $$

Only two primes under \(\sqrt{8191} = 91\) satisfy this equation. \(p = 53\) and \(p = 79\). So just need to check that that neither divide \(2^13 = 8191\) and neither does! \(\blacksquare\)

Fermat Primes

Recall that Fermat Primes are of the form \(2^n + 1\). Also recall that if \(n\) has an odd factor, then \(2^n+1\) is not prime. The reason for this is that \(x^{2n+1}+1\) is divisible by \(x+1\) since

$$ \begin{align*} x^{2n+1} + 1 = (x+1)(x^{2n} - x^{2n-1} + \cdots + 1) \end{align*} $$

In general if \(n = ab\) with \(a\) odd. Then,

$$ \begin{align*} 2^{n} + 1 = 2^{ab} + 1 = (2^b)^a + 1 \end{align*} $$

\((2^b)^a + 1\) is divisible by \(2^b + 1\). So for \(2^n + 1\) to be prime, \(n\) needs to be a power of 2. For example, the following are Fermat primes

$$ \begin{align*} 2^1 + 1 = 3, 2^2 + 1 = 5, 2^4 + 1 = 17, 2^8 + 1 = 257, 2^{16} + 1 = 65537 \end{align*} $$

So now we want check the following

Show that \(2^{2^n} + 1\) is prime.

As before, we can check all the primes up to \(\sqrt{2^{n} + 1}\). But we can use Fermat’s theorem to speed things up. How? Suppose \(p\) divides \(2^{2^n} + 1\). This just means that

$$ \begin{align*} 2^{2^n} + 1 &\equiv 0 \bmod p \\ 2^{2^n} &\equiv -1 \bmod p \end{align*} $$

Squaring both sides

$$ \begin{align*} 2^{2^{n+1}} &\equiv (-1)^2 \bmod p \\ 2^{2^{n+1}} &\equiv 1 \bmod p \end{align*} $$

So now, let \(d\) be the order of \(2 \mod p\). \(d\) must divide \(2^{n+1}\) so \(d\) must be some power of 2 less than \(2^{n+1}\) so

$$ \begin{align*} d \in \{1,2,4,\cdots,2^n,2^{n+1}\} \end{align*} $$

At the same time, \(d\) is the smallest power such that \(2^d \equiv 1 \bmod p\) and we also know that

$$ \begin{align*} 2^{2n} \equiv -1 \bmod p \end{align*} $$

So \(d\) can’t divide \(2^n\). Why? Suppose it did, then we can write \(2^n = d \cdot m\). Then

$$ \begin{align*} 2^{2^n} = 2^{d \cdot m} = (2^d)^m \end{align*} $$

But \(d\) is the order of 2 so

$$ \begin{align*} 2^{2^n} = (2^d)^m \equiv 1 \bmod p \end{align*} $$

This is a contradiction since \(2^{2^n} \equiv -1 \bmod p\). So \(d\) must be exactly \(2^{2^{n+1}}\). Also by Fermat, we know that

$$ \begin{align*} 2^{p-1} \equiv 1 \bmod p \end{align*} $$

This is because \((2,p) = 1\) and \(p\) is prime. Therefore, \(d = 2^{n+1}\) divides \(p-1\). This means that

$$ \begin{align*} p - 1 \equiv 0 \bmod 2^{n+1} \\ p \equiv 1 \bmod 2^{n+1} \end{align*} $$

This is a strong condition that the prime factors must satisfy. For example, take

$$ \begin{align*} 65537 = 2^{16} + 1 \end{align*} $$

If \(p \mid 65537\), then

$$ \begin{align*} p \equiv 1 \quad (\bmod 2^{4+1}=32) \end{align*} $$

and if

$$ \begin{align*} p \not\equiv 1 \bmod 32 \end{align*} $$

Then \(p \not\mid 65537\). So we only want to check any prime \(p\) less than \(\sqrt{65537} < 257\) such that \(p \equiv 1 \bmod 32\). If we look at any number that satisfies \(p \equiv 1 \bmod 32\), then we get

$$ \begin{align*} 33, 65, 97, 129, 161, 197,225 \end{align*} $$

Some of these are obviously not prime, we only need to check

$$ \begin{align*} 97, 197 \end{align*} $$

So now we can just use long division to verify that 65537 is prime. Neither of these divide 65537 so 65537 is prime.

Example

Is \(2^{2^5} + 1 = 2^{32} + 1\) prime?

Euler proved that this is not a prime. Using the condition we proved earlier, suppose that \(p \mid 2^{2^5} + 1\). This means that

$$ \begin{align*} p \equiv 1 \bmod 2^{5+1} = 64 \end{align*} $$

So the numbers that satisfy this are

$$ \begin{align*} 65, 129, 193, 257, 321, 385, 449, 513, 577, 641, .... \end{align*} $$

We can cross the numbers divisible by 5 or 3 and get

$$ \begin{align*} 129, 193, 257, 449, 577, 641, .... \end{align*} $$

What Euler discovered is that \(641 \mid 2^{2^5} + 1\). How do we check that 641 divides \(2^{2^5} + 1\)? An easy way is see that

$$ \begin{align*} 2^{16} = 154 \bmod 641 \end{align*} $$

So now

$$ \begin{align*} 2^{32} = (2^{16})^2 = (154)^2 \equiv -1 \bmod 641 \end{align*} $$

The idea is to take modulo 641 at every step to avoid large calculations.

Example

Suppose \(a^2 \equiv b^2 \bmod m\). Does this imply \(a \equiv \pm b \bmod m\)?

The answer is no. Suppose \(m = 8\). Then,

$$ \begin{align*} 1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \bmod 8 \end{align*} $$

But

$$ \begin{align*} 1 \not\equiv \pm 3 \bmod 8 \end{align*} $$

It does work if \(m\) is prime. Why? Observe that

$$ \begin{align*} a^2 \equiv b^2 \bmod p \\ a^2 - b^2 \equiv 0 \bmod p \\ (a-b)(a+b) \equiv 0 \bmod p \end{align*} $$

Because \(p\) is prime, then either \((a-b)\equiv 0 \bmod p\) or \((a+b) \equiv 0 \bmod p\).

Example

Suppose \(a^2 \equiv b^2 \bmod m^2\). Does this imply \(a \equiv \pm b \bmod m\)?

The answer is still no. Take primes \(p\) and \(q\) and let \(m = pq\). Now suppose

$$ \begin{align*} a &\equiv b \bmod p^2 \\ a &\equiv -b \bmod q^2 \end{align*} $$

But these two imply that

$$ \begin{align*} a^2 &\equiv b^2 \bmod p^2 \\ a^2 &\equiv (-b)^2=b^2 \bmod q^2 \end{align*} $$

We can combine both to get

$$ \begin{align*} a^2 \equiv b^2 \bmod p^2 \cdot q^2 \\ a^2 \equiv b^2 \bmod m^2 \end{align*} $$

The claim is that

$$ \begin{align*} a \not\equiv \pm b \bmod m \end{align*} $$

Why? suppose for the sake of contradiction that \(a \equiv \pm b \bmod m=pq\). Specifically suppose that

$$ \begin{align*} a &\equiv -b \bmod pq \end{align*} $$

Since \((p,q) = 1\), then

$$ \begin{align*} a &\equiv -b \bmod p \\ \end{align*} $$

but now recall that we also have \(a \equiv b \bmod p^2\) so

$$ \begin{align*} a &\equiv b \bmod p \end{align*} $$

This is a contradiction. We can’t have both \(a \equiv b \bmod p\) and \(a \equiv -b \bmod p\) at the same time. Similarly, we can arrive at a contradiction if \(a \equiv b \bmod pq\) by using the other assumption that \(a \equiv -b \bmod q^2\). \(\ \blacksquare\)

References