Prove that if $p \equiv 7 \pmod{8}$, the solutions to the congruence $x^2-34x+1 \equiv 0 \pmod{p}$ are both quadratic residues modulo p.

Question

Consider the quadratic congruence \begin{equation} x^2-34x+1 \equiv 0 \pmod{p}, \end{equation} where $p \equiv 7 \pmod{8}$. There are two solutions, since we can write it as \begin{equation} (x-17)^2 \equiv 288 \pmod{p} \end{equation} and check that $288$ is a quadratic residue in this case. Since the product of solutions is congruent to $1 \pmod{p}$, then both of them are either quadratic residues or quadratic non-residues. I have checked multiple primes, and it appears that both solutions are always quadratic residues. I'm not sure how to prove it; I feel like I'm missing something obvious. Any hint would be much appreciated!

Answer 1

When $p\equiv7\pmod8$, then $y^2=2$ has solutions $\pmod p$,

which I will write as $y=\pm\sqrt2\pmod p$.

The solutions of the congruence $x^2-34x+1\equiv0\pmod p$

are seen to be $x=17\pm12\sqrt2\pmod p,$ using the quadratic formula.

Can you show that $x=(3\pm2\sqrt2)^2$?

Answer 2

Easy way: $ $ applying my denesting formula we can compute $\color{#0a0}{\sqrt x}\,$ by completing a square

$\qquad \qquad\ 34x = x^2+1 \iff 36x = (x\!+\!1)^2\!\!\iff x = \left(\!\color{#0a0}{\dfrac{x\!+\!1}6}\!\right)^{\!2}$

Remark $ $ So $\:\!x\:\!$ has a $\:\!\rm\color{#0a0}{square\ root}\:\!$ in any ring where $\color{#0a0}{1/6}\,$ exists, e.g. $\:\!\Bbb Z_n\:\!$ for all $\:\!n\:\!$ coprime to $\:\!6,\,$ e.g. OP's $\,n=$ prime $p\equiv 7\pmod{\!8},\,$ where $\,p\neq 2,3\Rightarrow p\,$ coprime to $\:\!6$.