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Show that prime factors of $n^2+1$ are congruent to $1 \bmod 4$

Solution:

I think that if I denote $g$ as a primitive root, I know that the order of $g=p-1$

In other words, since $g^{p-1}\equiv 1\bmod p $ and $p$ is odd, I know that $(p-1)/2$ exists. Now we know that $g^{(p-1)/2} \equiv -1 \bmod p$. So $-1$ will infact be a quadratic residue if $g^{(p-1)/4}$ exists.

Since $(p-1)/4$ can be any number from the set ${1,2,3,...,p-1}$ I can conclude that the number I'm looking for is of the form $p=4k+1$

The product of any primes of the form $4k+1$ is also of the form $4m+1$.

Is this proof correct? It seems to me that I magically land on the $-1$ congruence out of the blue. I think what I'm missing is to use somehow that $n^2+1$ factors always contain $-1$ as a residue or something.

GRS
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  • "The product of any primes of the form $4k+1$ is also a prime of the form $4m+1$". Are you sure that's true? –  Nov 06 '15 at 18:41
  • Sorry, I meant that it is also of the same form. Not necessarily a prime. $(4k+1)(4n+1)=4(4kn+k+n)+1$ – GRS Nov 06 '15 at 18:46
  • Then that's exactly right - and you don't necessarily need to do the math because $1*1 \equiv 1 \mod 4$. –  Nov 06 '15 at 18:52
  • There are many ways to prove this. See here for more ways. – user236182 Nov 06 '15 at 21:28

1 Answers1

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You kinda have the right idea, but explained it badly.

If $p |n^2+1$ you get that $$n^2 \equiv -1 \pmod{p}$$

This shows that $-1$ is a quadratic residue modulo $p$. If you are familiar with Legendre symbol just calculate $\left( \frac{-1}{p} \right )$ otherwise, it is kinda the idea you tried:

Write $n =g^a$ where $g$ is a primitive root and $0 \leq a < p-1$. Then $$n^2=-1$$ implies $$g^{2a} =-1 =g^{\frac{p-1}{2}}$$ implies that $p-1| 2a-\frac{p-1}{2}$. Use this and $0 \leq a < p-1$ to deduce that $2a=\frac{p-1}{2}$.

N. S.
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