MHB Problem of the Week #98 - April 14th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Let $p$ be a prime. Prove that the polynomial $x^4+1$ splits mod $p$ either into two irreducible quadratics or into four linear factors.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Opalg. You can find his solution below.

[sp]The key step here seems to be Gauss's law of quadratic reciprocity (see here):

If $p\equiv1 \pmod4$, the negative of a residue modulo p is a residue and the negative of a nonresidue is a nonresidue.
If $p\equiv3 \pmod4$, the negative of a residue modulo p is a nonresidue and the negative of a nonresidue is a residue.​

Case 1: $p=2$. In this case $x^4+1 \equiv (x+1)^4$, with four linear factors.

Case 2: $p\equiv1 \pmod4$. Since $1$ is obviously a quadratic residue, it follows from Gauss's law that so is $-1$, say $-1 = c^2$. Then $x^4+1 = (x^2+c)(x^2-c)$. Again by Gauss's law, either $c$ and $-c$ are both quadratic residues or neither of them is. So either both the quadratic factors are irreducible, or they both factorise giving four linear factors.

Case 2: $p\equiv3 \pmod4$. The factorisation $x^4+1 = (x^2 + ax + b)(x^2 - ax +b)$ holds, with $b = \pm1$, provided that $a^2 = 2b = \pm2$. But by Gauss's law either $2$ or $-2$ is a quadratic residue. So a solution for $a$ and $b$ exists. Also, $x^2 + ax + b$ has a factorisation $x^2 + ax + b = (x- c)(x-d)$ if and only if $x^2 - ax + b = (x+ c)(x+d)$. So either both the quadratic factors are irreducible, or they both factorise giving four linear factors.[/sp]