solvable congruent

**alexandrabel90** · April 20th 2012, 07:48 AM

How do you proof that -1 congruent to x^4 mod p is solvable iff (-1)^(p-1/d) is congruent to 1 mod p where d= gcd(4,p-1)?

i was trying to use euler theorem that (-1)^(p-1) is congruent to 1 mod p then how else can i continue?

**Sylvia104** · April 20th 2012, 06:39 PM

I take it you want $p$ to be prime. The case $p=2$ is trivial, so we'll assume $p$ to be odd. Then $d=2$ if $p\equiv3\mod4$ and $d=4$ if $p\equiv1\mod 4.$

If $p\equiv3\mod4$ then by Euler's criterion $\left(\frac{-1}p\right)=(-1)^{\frac{p-1}2}=-1.$ In this case $x$ is not a quadratic residue modulo $p$ (and so it certainly can't be a quartic residue modulo $p);$ we also have $(-1)^{\frac{p-1}d}=(-1)^{\frac{p-1}2}=-1\not\equiv1\mod p.$

Consider $p\equiv1\mod4.$

(i) Suppose $x^4\equiv-1\mod p.$ As $d=4$ in this case, $-1\equiv x^d\mod p$ $\implies$ $(-1)^{\frac{p-1}d}\equiv x^{p-1}\mod p\equiv1\mod p$ by Fermat's little theorem.

(ii) Conversely suppose $(-1)^{\frac{p-1}d}=(-1)^{\frac{p-1}4}\equiv1\mod p.$ Then $\frac{p-1}4$ is even, i.e. $8\mid p-1.$ Let $k$ be a primitive root $\mod p$ and set $x=k^{\frac{p-1}8}.$ Then $x^8\equiv1\mod p$ $\implies$ $p$ divides $x^8-1=\left(x^4+1\right)\left(x^4-1\right).$ But $p$ cannot divide $x^4-1$ otherwise $k^{\frac{p-1}2}\equiv1\mod p$ which would contradict the fact that, as a primitive root, $k$ has order $p-1$ in the multiplicative group of the integers modulo $p.$ Hence $p$ divides $x^4+1$ and we are done.

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