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I'm looking at this funny little problem involving proving the existence of an infinite number of primes of a certain form:

Prove that there are infinitely many prime numbers expressible in the form $8n+1$ where $n$ is a positive integer.

My proof goes like this:

Assume by way of contradiction that there are only finitely many of such prime numbers, say $p_1,p_2,\ldots,p_r$. Consider $p = 16p_1^4p_2^4\cdots p_r^4 +1 = (2p_1p_2 \cdots p_r)^4+1$. Recall that the congruence $x^4 \equiv -1 \mod p$ is solvable if and only if $p \equiv 1 \mod 8$. $p$ cannot be a prime of the form $8n+1$, but $p \equiv 1 \mod 8$, contradiction. Thus, there must be infinitely many prime numbers of the form $8n+1$.

I'm not quite sure if I actually achieve a contradiction here. Can you help me clarify this?

Martin Sleziak
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Numbersandsoon
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2 Answers2

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This idea seems fine, although notice that $p$ does not have to be your number itself, but can be one of its divisors.

$16p_1^4p_2^4\ldots p_r^4+1$ must have some prime divisor $p$, and it is not one of the existing $p_i$, nor is it 2. Moreover $x=2p_1p_2\ldots p_r$ satisfies the congruence $x^4+1\equiv 0 \pmod p$, and therefore $p\equiv 1\pmod 8$, a contradiction.

user7530
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  • Incidentally, is there a clever way of seeing that last fact? Certainly you can argue that the order of $x$ must be 8, and therefore $8\vert \phi(p)$, but that requires rather a lot of machinery for a very elementary problem. – user7530 Dec 11 '13 at 06:52
  • You consider the existence of infinitely many primes in an arithmetic progression more elementary than these facts about the orders of elements modulo $p$? – Greg Martin Dec 11 '13 at 08:13
  • This particular arithmetic progression, yes. – user7530 Dec 11 '13 at 14:43
  • A chacun son goût! – Greg Martin Dec 11 '13 at 18:32
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Claim:

The odd prime divisors of $n^4+1$ are of the form $8k+1$

From $n^4\equiv -1\pmod{p}$ we have $n^8\equiv 1\pmod{p}$, therefore the order of $n$ modulo $p$ is $8$, and from the following theorem:

If $a$ is of order $k$ modulo $n$, then $a^h\equiv 1\pmod{n}$ iff $k|h$. Also $k|\phi(n)$.

It follows that $8|\phi(p)$, or in other words: $p=8k+1$.

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Salech Alhasov
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    This is neat, but does not yet prove infiniteness. What if the set of all values of $n^4 + 1$ has only a few odd prime divisors? – darij grinberg Oct 13 '21 at 05:12
  • @darijgrinberg Let $n_1=2^4+1$. Suppose we have chosen integers $n_1<...<n_k$ such that $n_j$ is coprime to $n_1,...,n_{j-1}$ and is $\equiv1\pmod8$. Let $n_{k+1}=\left(2\prod_{i=1}^k n_i\right)^4+1$. Then $n_{k+1}>n_{k}$ is coprime to $n_1,...,n_{k}$ and $\equiv1\pmod8$. We have an injective sequence ${n_k}_{k=1}^{\infty}$. Each integer $n_k$ is $>1$ and is odd and thus must have a prime divisor $p_k$. Since the $n_k$'s are pairwise coprime, the $p_k$'s are pairwise coprime. For each $k$, since $n_k$ is of the form $n^4+1$, $p_k\equiv1\pmod8$. (This is essentially user7530's answer) – 19021605 Dec 16 '25 at 08:29