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.

  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ No, I do not. @BillDubuque $\endgroup$ – virreand Jan 17 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.