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I was reading The Higher Arithmetic by H. Davenport which was proving that there are infinitely many primes of both the forms $4x+1$ and $4x-1$.

The proof for $4x-1$ was trivial:
Let $N=4(x_1,x_2,…,x_n)-1$
This itself is a number of the form $4x-1$. There must be a factor of $N$ of the form $4x-1$ which cannot be any of the primes $x_1,x_2,…,x_n$, therefore there exists a prime in this series which is different from any $x_1,x_2,…,x_n$, therefore proving the proposition.

The same logic, however, doesn’t apply to $4x+1$ as it doesn’t follow that this number will necessarily have a prime factor of such form.

It then goes on to explain that in this case we can let $M=(y_1,y_2,…,y_n)^2+1$ which must be entirely composed of prime factors of the form $4x+1$ together with possibly $2$. Since $M$ is not divisible by any of the primes $y_1,y_2,…,y_n$, there must be infinitely many primes of the form $4x+1$.

My confusion lies where it states that any number of the form $a^2+1$ must be entirely composed of prime factors of the form $4x+1$ along with possibly $2$. No proof of this statement is provided and I was wondering if someone could give one (preferably using high school mathematics as I am a high school student myself).

Bill Dubuque
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    If a prime $p$ divides an expression of the form $n^2+1$ that means $-1$ is a quadratic residue $\bmod p$. And that happens when $p$ is $2$ or congruent to $1\bmod 4$. – Asinomás Apr 17 '22 at 20:49

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As Asinomas pointed out in the comments, if a prime $p \mid a^2 +1$, then we know there exists an integer $x$ such that $x^2 \equiv -1$ (mod $p$). In other words, $-1$ is a quadratic residue mod $p$. This only occurs when $p=2$ or $p\equiv 1$ (mod $4$). If you are unsure of why this is the case, check out this post.

Clyde Kertzer
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