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Given an odd prime $p$, I want to prove that $x^4 \equiv -4$ (mod $p$) is solvable if and only if $p \equiv 1$ (mod $4$). More specifically, I want to prove this using a hint, which says to first factorize $x^4+4$.

I proved that whenever the congruence is solvable, we have $p \equiv 1$ (mod $4$), however, I didn't manage to do that using the factorization but rather using Legendre symbols $(\cdot / \cdot)$. But I would like to prove the converse using the factorization.

This is the way I have tried to solve it but I'm unsure of whether it is correct or not:

So first we factorize $x^4+4$ as $$x^4 + 4 = (x^2 + 2x + 2)(x^2 - 2x + 2) = ((x+1)^2 + 1)((x-1)^2 + 1), $$ which can be written as $$ = (a^2 + 1)(b^2 + 1). $$

Now we want to show that there exists $a, b$ such that $(a^2 + 1)(b^2 + 1) \equiv 0$ (mod $p$). So in modulo $p$, we have that $$a^2 + 1 \equiv 0 \iff a^2 \equiv -1 $$ has a solution iff $(-1/p) = 1$. We have that $(-1/p) = (-1)^{(p-1)/2} = 1$ since $(p-1)/2$ is even whenever $p \equiv 1$ (mod $4$). The same goes for $b^2 + 1$, and we are done.

Was this correct? If not, I could use some advice on how to solve this.

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    $\begingroup$ If p is an odd prime, (-1/p)=1 iff p=1 mod 4. Your prove is correct and gives you also the converse. $\endgroup$ – Javier Linares Jan 16 at 13:51
  • $\begingroup$ Thank you for checking it! $\endgroup$ – virreand Jan 16 at 14:03
  • $\begingroup$ Is not this competent with Fermat theorem: $p=a^2+b^2$ iff $p=4k+1$? $\endgroup$ – sirous Jan 17 at 18:27
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The factorization you have found shows that $x^4 \equiv -4$ has a solution iff $y^2 \equiv -1$ has a solution: $$ x^4 + 4 = ((x+1)^2 + 1)((x-1)^2 + 1) $$

Now $y^2 \equiv -1$ has a solution iff there is an element of order $4$ mod $p$. This happens iff $4$ divides $p-1$ because the group of units mod $p$ is cyclic.

So you don't need Legendre symbols.

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  • $\begingroup$ Thank you for your answer. I have not learned about cyclic groups yet, so I will have to come back to this at a later time. $\endgroup$ – virreand Jan 16 at 16:38
  • $\begingroup$ @virreand Do you understand the claimed equivalence in the first sentence? At lhf: How do you propose to prove that (unjustified) claim? At this level one should not leave such claims unjustified. $\endgroup$ – Bill Dubuque Jan 17 at 1:06
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    $\begingroup$ No, I do not. @BillDubuque $\endgroup$ – virreand Jan 17 at 17:55

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