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prove that for all primes p, the roots of $(x^2+1)(x^4-4) \in (\mathbb{Z}/p\mathbb{Z})[x]$ have roots in $(\mathbb{Z}/p\mathbb{Z})$.

I thought that if this polynomial has roots, we have two cases:

$x^2 + 1 = 0$ or $x^4 - 4 = 0$.

I guess a root in $(\mathbb{Z}/p\mathbb{Z})$ is of the form $0 \pmod{p}$.

So i think that we need to prove that for all primes:

$x^2 \equiv -1 \pmod{p}$ has a solution and:

$x^4 \equiv 4 \pmod{p}$.

For the first equation we see $\left(\frac{-1}{p}\right) = 1 \iff p \equiv 1 \pmod {4}$.

For the second equuation we see:

$x^4 \equiv 4 \pmod{p} \implies x^2 \equiv 2 \pmod{p}$ or $x^2 \equiv -2 \pmod{p}$.

thus we see $\left(\frac{2}{p}\right) = 1 \iff p \equiv +- 1 \pmod 8$ and :

$\left(\frac{-2}{p}\right) = \left(\frac{2}{p}\right) \left(\frac{-1}{p}\right) = 1 \iff p \equiv 1 \pmod 4$ or $p \equiv 3 \pmod{8}$.

Combining these together we get all primes in the form: $1,3,7 \pmod{8}$ will have roots. but how about $2$ and the primes of the form $5 \pmod{8}$.

Maybe i am doing this completely wrong, so could someone hint me in the right direction :)

Kees

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    $\begingroup$ $\left(\frac{-2}{p}\right)=1\iff p\equiv \{1,3\}\pmod{8}$, so you were wrong there. It should be:$$\left(\frac{2}{p}\right) \left(\frac{-1}{p}\right) = 1 \iff \begin{cases}\begin{cases}p \equiv 1 \pmod 4\\p\equiv \pm 1\pmod{8}\end{cases}\\\text{or}\\\begin{cases}p\equiv 3\pmod{4}\\p\equiv \{3,5\}\pmod{8}\end{cases}\end{cases}$$ which is $\iff p\equiv \{1,3\}\pmod{8}$. $\endgroup$ – user236182 Oct 30 '15 at 20:41
  • $\begingroup$ $x^4\equiv 4\pmod{p}$ should be $\iff$, not just $\implies$, $x^2\equiv \pm 2\pmod{p}$. You must have equivalences everywhere. $\endgroup$ – user236182 Oct 30 '15 at 20:43
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You have some mistakes in your work that I explained in the comments. Here's how it should be:

So it's equivalent to proving at least one of $x^2\equiv -1\pmod{p}$ or $x^4\equiv 4\pmod{p}$ is solvable, no matter what prime $p$ you choose.

$1)\ \ \ $ $x^2\equiv -1\pmod{p}$ is solvable iff either $p=2$ or $\left(\frac{-1}{p}\right)=1$, i.e. iff either $p=2$ or $p\equiv 1\pmod{4}$.

$2)\ \ \ $ $x^4\equiv 4\pmod{p}\iff x^2\equiv \pm 2\pmod{p}$. We know $x^2\equiv 2\pmod{p}$ is solvable iff either $p=2$ or $p\equiv \pm 1\pmod{8}$ and $x^2\equiv -2\pmod{p}$ is solvable iff either $p=2$ or $p\equiv \{1,3\}\pmod{p}$ (see the comments). Therefore $x^4\equiv 4\pmod{p}$ is solvable iff either $p=2$ or $p\equiv \{1,3,7\}\pmod{8}$.

$1)$ implies $\left(x^2+1\right)\left(x^4-4\right)\equiv 0\pmod{p}$ is solvable if $p=2$ or $p\equiv \{1,5\}\pmod{8}$ and $2)$ implies it's solvable if $p\equiv \{3,7\}\pmod{8}$. We conclude it's solvable for all primes.

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  • $\begingroup$ how do you see these equations are always solvable for $p=2$. Do you just pick $x=1$ or $x= 2$? $\endgroup$ – Kees Til Oct 30 '15 at 21:07
  • $\begingroup$ @KeesTil $x^2\equiv -1\pmod{2}$ is solvable because $x\equiv 1\pmod{2}$ is a solution. Also $x^2\equiv \pm 2\pmod{2}$ is solvable because $x\equiv 0\pmod{2}$ is a solution. $\endgroup$ – user236182 Oct 30 '15 at 21:34
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Primes that are 5 mod 8 are necessarily 1 mod 4, so you already have those. As for 2, just observe that $x^2 + 1 = (x+1)^2$ mod 2.

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  • $\begingroup$ Note OP's work has a mistake. $\endgroup$ – user236182 Oct 30 '15 at 20:44
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Hint The primes of the form $5 \pmod{8}$ are of the form $1 \pmod{4}$.

Also, for $p=2$ you have $$x^2+1\equiv X^2-1 \pmod{2}$$

For problems like this, the question can be solved by hand for $p=2$.

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  • $\begingroup$ Note OP's work has a mistake. $\endgroup$ – user236182 Oct 30 '15 at 20:44

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