Please check my proof.

I know this proof can be easier if use legendre symbol, but the problem does not allow to use it.

Suppose $a$ is quadratic residue from Euler's criteria

$a^{(p-1)/2}\equiv 1 \pmod{p}$

and $b$ is quadratic nonresidue It will be $b^{(p-1)/2}\not\equiv 1 \pmod{p}$ or $b^{(p-1)/2}\equiv c \pmod{p}$ for some c then

$(ab)^{(p-1)/2}\equiv c \pmod{p}$ or

$(ab)^{(p-1)/2}\not\equiv 1 \pmod{p}$ by congruence's property

If $a\equiv b \pmod{m}$ and $c\equiv d \pmod{m}$ then $ac\equiv bd \pmod{m}$ then the product of quadratic residue and quadratic nonresidue is quadratic nonresidue


Yes, this proof is correct, and indeed is the shortest proof (in my opinion).

Note that if $b^{(p-1)/2}\not\equiv1\pmod p$, then necessarily $b^{(p-1)/2}\equiv-1\pmod p$ (that is, we have to have $c=-1$). This is because $(b^{(p-1)/2})^2\equiv1\pmod p$ by Fermat's little theorem, and the only solutions to $x^2\equiv1\pmod p$ are $x\equiv\pm1\pmod p$.


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