Let $x$ be an integer and let $d$ be a factor of $2x^2 − 1$. Prove that $d^2 ≡ 1\bmod 16$

Question

So I tried expanding and got $d^2$ to be $4x^4-4x^2+1$. From what I think, I believe that I have to prove that $4x^4-4x^2 ≡ 0\mod 16$, but I have no idea on how I can do that and my approach could be wrong.

Any help would be much appreciated.

Answer 1

If $d$ is a factor of $2x^2-1$, then any prime factor $p$ of $d$ is a factor of $2x^2-1$.

Prime $p>2$ is a factor of $2x^2-1\iff 2$ is a quadratic residue modulo $p$

$\iff p\equiv\pm1\bmod 8.$ Obviously $2$ is not a factor of $2x^2-1$.

So $d$ is a factor of $2x^2-1\iff d\equiv\pm1\bmod8 \iff d^2\equiv1\bmod16$.

The last equivalence follows because if $8|d\pm1$ then $16|(d+1)(d-1)=d^2-1$.