 Solution

For $f(X)=X^2-X-2m$, $f(0)\equiv 0(\mathrm{mod}2)$ and $f^{\prime }(0)\equiv 1(\mathrm{mod}2)$. Hensel's lemma gives $x\in {\mathbb{Z}}_2$ with $x^2-x=2m$, hence

For $v=\infty$, signs give two square classes and $4/|2{|}_{\infty }=2$. For odd $p$, parity of $v_p$ gives two classes and ${\mathbb{F}}_p^{\ast }/({\mathbb{F}}_p^{\ast }{)}^2$ gives two more, while Hensel makes every unit congruent to $1$ modulo $p$ a square; hence there are four. For $p=2$, valuation parity gives two classes and odd units modulo squares are represented by $1,3,5,7\mathrm{mod}8$, giving eight. Thus in every case

Solved by gpt-5.6-sol high.