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This question is from my problem set in number theory. I tried it a week ago and again today but both times I couldn't solve it.

Problem: If a prime p divides an integer of the form $n^2+1$ , then $p\not\equiv 3 \pmod{4}$.

I thought of attempting this problem by assuming $p\equiv 3\pmod{4} $. But I am unable to move towards anything concrete.

Can you please give some hints?

Thanks!

Arturo Magidin
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    Hint: if $p$ divides $n^2+1$, then $x^2\equiv -1\pmod{p}$ has a solution. – Arturo Magidin Apr 26 '22 at 17:54
  • @ArturoMagidin, Euler's criterion : $a^{\frac {p-1}2}\equiv -1\bmod p$ , if $x\not\equiv m^2\bmod p$. $n^2\equiv -1\bmod p$ means that is you can assume $\frac{p-1}2=2$, which gives $p=5$ and $5\equiv 1\bmod 4 \not\equiv 3 \bmod 4$. – sirous Apr 27 '22 at 00:19
  • @sirous: I cannot understand what you are saying, nor do I see how "you can assume[...] $p=5$". Euler's criterion says that for odd prime $p$, $x^2\equiv a\pmod{p}$ has solutions (for $\gcd(p,a)=1$) if $a^{(p-1)/2}\equiv 1\pmod{p}$, and has no solutions if $a^{(p-1)/2}\equiv -1\pmod{p}$. Here, $a=-1$. So $x^2\equiv -1\pmod{p}$ has a solution if and only if $(p-1)/2$ is even. – Arturo Magidin Apr 27 '22 at 00:28
  • @ArturoMagidin, I just compared relations and concluded $\frac{p-1}2=2\Rightarrow p=5$. – sirous Apr 27 '22 at 00:32
  • @sirous And I don't know where you get $(p-1)/2=2$. There is no reason to think that $(p-1)/2=2$ just because $(-1)^{(p-1)/2}\equiv 1\pmod{p}$. Any even number will give you $(-1)^{(p-1)/2}\equiv 1\pmod{p}$, not just $2$. Why not $6$ (when $p=13$)? Why not $8$? $36$? $504$? – Arturo Magidin Apr 27 '22 at 00:37
  • I don't understand the reason behind so many downvotes –  Apr 27 '22 at 06:23
  • Does this answer your question? Which primes $p$ satisfy $n^2 \equiv -1 \mod p$ for a perfect square $n^2$? – 311411 Apr 27 '22 at 13:16
  • Comment:

    An experimental approach:

    Due to a hypothesis by Schintzel for any natural number $s$ , there are infinitely many natural numbers like $n$ such that $n^2+1$ is product of $s$ prime numbers. For example:

    $13^2+1=2\cdot 5\cdot 17$

    $17^2+1=2\cdot 5\cdot 29$

    $23^2+1= 2\cdot 5\cdot 53$

    As you see all odd prime factors are ($1\bmod 4$).Also in all examples ($a\equiv \pm 1 mod 4$). This can be considered as the condition for primality of a.

     – sirous Apr 27 '22 at 14:40
  • This is in fact special case of Euler's equation $4xy-x=a^2+n$ where $n=1$ which has no solution. Goldbach proved it. Rewrite $4xy-x=x(4y-1)$. Suppose $p=4y-1$ or $p\equiv -1 \bmod 4\equiv 3\bmod 4$, that is p can not be ($3\bmod 4$) due to proof of Golbach. – sirous Apr 27 '22 at 19:04

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You're right on assuming that $p\equiv 3\pmod{4}$ and then trying arrive at a contradiction. Here's how it may work out.

Suppose that $p\equiv 3\pmod{4}$ and $p$ divides $n^2+1.$ If $p$ divides $n^2+1,$ then we have; $$n^2+1\equiv 0\pmod{p}$$ $$n^2\equiv -1\pmod{p}.$$

Now, since we have assumed that $p\equiv 3\pmod{4}$ we know that $p=4k+3$ for some positive $k.$ Thus, we have; \begin{align*} n^{p} &\equiv n^{4k+3}\pmod{p}\\ &\equiv n\left(n^{4k+2}\right)\pmod{p}\\ &\equiv n\left(n^2\right)^{2k+1}\pmod{p}. \end{align*}

Note that $n^2\equiv -1\pmod{p},$ so we have; \begin{align*} n^{p} &\equiv n\left(n^2\right)^{2k+1}\pmod{p}\\ &\equiv n\left(-1\right)^{2k+1}\pmod{p}\\ &\equiv n(-1)\pmod{p}\\ &\equiv -n\pmod{p}. \end{align*}

But by the Fermat's Little Theorem, we have $n^p\equiv n\pmod{p}.$ So we have $n\equiv -n\pmod{p},$ which simplifies to $p\mid 2n.$ However, we can see that this is incorrect as $p$ also divides $n^2+1$ which is coprime to $2n$. So we have finally arrived a contradiction and your claim is proven.

MathMinded
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    +1. My edit was for typos in line 6 where you had n for p. – DanielWainfleet May 08 '22 at 05:59
  • In case any reader is wondering why $n\equiv -n\mod p$ is contradictory, it implies $2n\equiv 0 \mod p$, implying $0\equiv 4(n^2+1)\equiv (2n)^2+4\equiv 4\mod p, $ implying $p$ divides $ 4.$ – DanielWainfleet May 08 '22 at 06:09
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    @DanielWainfleet Thanks for editing! Let me just add the explanation of why $n\equiv n\pmod{p}$ is contradictory to my answer. – MathMinded May 08 '22 at 15:01