Proof: For any prime number $p$, there must be integer $a$, $b$, $c$, $d$ such that $x^4+1 \equiv (x^2+ax+b)(x^2+cx+d)\ (mod\ p)$.

Asked Jan 13 '19 at 14:16

Active Jan 13 '19 at 14:16

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This is a question comes from a book about Number theory, and the question is related to Quadratic residue.

I start with

$$ x^4+1 \equiv (x^2+ax+b)(x^2+cx+d) \\ \equiv x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd\ (mod\ p) $$

and then

$$ \begin{cases} a+c \equiv b+d+ac \equiv ad+bc \equiv 0\ (mod\ p)\\ bd \equiv 1\ (mod\ p) \end{cases} $$

and I don't know what to do next.

The Legendre symbol and Jacobi symbol Properties and Calculating is ok for me.

asked Jan 13 '19 at 14:16

HenryLiu

1

Hint: If $-1$ is a quadratic residue, then $y^2+1$ factors, and hence so does $x^4+1$. If not, then exactly one of $2$ and $-2$ is a quadratic residue, so that $4$ is a $4$-th power mod $p$. Thus factoring $x^4+1$ is equivalent to factoring $x^4+4$. The latter has been done by Sophie Germain. – darij grinberg Jan 13 '19 at 14:24

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