Infinitely Many Primes With Last Digit 1

Before discussing Euler’s theorem, recall Fermat’s Theorem from the lecture 09

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

We will use it to show that there are infinitely many primes with last digit equal to 1. For example

$$ \begin{align*} 11, 31, 61, 71, 101, \cdots \end{align*} $$

We can show that this sequences goes on forever. This is in fact a special case of Dirichlet’s theorem which says if we have an arithmetic progression, then there are infinitely many primes in that progression. Here, we will focus on the special case of primes of the from \(10m+1\). First, it is easy to show that there are infinitely many primes of the form \(5n+1\) because \(5\) is a prime here.

There are infinitely many primes of the form \(p = qn+1\) where \(q\) is prime.

Proof: let \(N = a^q - 1\) where \(q\) is prime. We’re going to analyze the prime factors that divide \(N\). So suppose \(p \mid N\) where \(p\) is prime. Since \(p\) divides \(q^q - 1\), then

$$ \begin{align*} a^{q}-1 \equiv 0 \pmod{p} \\ a^{q} \equiv 1 \pmod{p} \end{align*} $$

Let the order of \(a \pmod{p}\) be \(o(a)\). By lecture 10, \(o(a)\) must divide \(q\). But \(q\) is prime, so \(o(a)\) either 1 or \(q\).
Case 1: \(o(a) = 1\). Then

$$ \begin{align*} a^{1} &\equiv 1 \pmod{p} \end{align*} $$

Case 2: The order is \(q\). By Fermat we know that \(a^{p-1} \equiv 1 \pmod{p}\). This implies that \(q\) must divide \(p-1\). Therefore

$$ \begin{align*} p &\equiv 1 \pmod{p} \end{align*} $$

This condition almost gets us the primes of the form that we want. Why? because this implies that

$$ \begin{align*} p - 1 &= qn \\ p &= 1 + qn \\ \end{align*} $$

However, condition one though allows primes \(p\) such that \(p \mid a - 1\), so they are not necessarily of the form that we want so we must eliminate case 1. To do so, instead of defining \(N = a^q - 1\), let \(N\) be

$$ \begin{align*} N = \frac{a^q - 1}{a - 1} \end{align*} $$

and assume that \(p \mid \frac{a^q - 1}{a - 1}\). This this also means that \(p \mid a^q - 1\). Therefore, we will have the same cases for the order of \(a \pmod{p}\). Recall that in case 1, we had \(a \equiv 1 \pmod{p}\). Also observe that

$$ \begin{align*} \frac{a^q - 1}{a - 1} = a^{q-1} + a^{q-2} + \cdots + 1 \end{align*} $$

Taking \(\bmod p\), then

$$ \begin{align*} \frac{a^q - 1}{a-1} \pmod{p} = (a^{q-1} + a^{q-2} + \cdots + 1) \pmod{p} \end{align*} $$

But since \(a \equiv 1 \pmod{p}\), then

$$ \begin{align*} \frac{a^q - 1}{a-1} \bmod p &= (1 + 1 + ... 1) \pmod{p} \\ &= q \pmod{p} \end{align*} $$

But also by assumption \(p \mid \frac{a^q - 1}{a - 1}\) so \(\frac{a^q - 1}{a - 1} \equiv 0 \pmod{p}\). So combining the following

$$ \begin{align*} \frac{a^q - 1}{a-1} &\equiv q \pmod{p} \\ \frac{a^q - 1}{a-1} &\equiv 0 \pmod{p} \\ \end{align*} $$

But this just means that \(q \equiv 0 \pmod{p}\) so \(q \mid p\). But \(q\) and \(p\) are both primes. This means that we must

$$ \begin{align*} p = q \\ \end{align*} $$

Case 2: we saw showed that \(p \equiv 1 \mod q\) which meant that \(p\) has the form

$$ \begin{align*} p = 1 + qn, \quad n \in \mathbb{Z} \end{align*} $$

Finally, for integers \(a \geq 2\). As \(a\) increases, \(N\) grows without bound, so the set of all prime divisors of such \(N\) is infinite. We saw that any prime \(p \mid N\) must satisfy either:

  • \(p = q\)
  • \(p \equiv 1 \pmod{q}\)

Since only finitely many of the \(N\) can be divisible by the fixed prime \(q\), it follows that infinitely many of their prime divisors must satisfy

$$ \begin{align*} p \equiv 1 \pmod{q}. \end{align*} $$

Therefore, there are infinitely many primes of the form \(p = qn + 1\) for some integer \(n\). \(\ \blacksquare\)

Example

Let’s fix \(q = 5\) and let’s see how this will work. So now \(N\) will be

$$ \begin{align*} N = \frac{x^5 - 1}{x - 1} = x^4 + x^3 + x^2 + x + 1 \end{align*} $$

So now suppose \(p \mid N\).

Euler’s Theorem is a generalization of Fermat’s Theorem

References