If $F(x,y)$ is a polynomial which is not a square, then how often is the specialization $F(x,a)$ a square? Suppose $F(x,y)$ is a polynomial in two variables over a field $K$, and $F(x,y)$ is not a square. When is $F(x,a)$ a square for $a\in K$? I would guess that Hilbert's Irreducibility Theorem might help (if $K$ is hilbertian), but I am not sure how.
 A: Write $F(x,y)=G(x,y)^2H(x,y)$ with polynomials $G,H$, such that $H(x,y)$, considered as a polynomial in $x$ over $K(y)$, is squarefree. If $H(x,y)$ does not involve $x$, so $H(x,y)=h(y)$, then you are reduced to an arithmetical question depending on $K$. For instance if $K$ is Hilbertian, and $h(y)$ is a not a square, then $h(a)$ is not a square infinitely often.
If however $H(x,y)$ involves $x$, then $H(x,y)$ and its derivative $H'(x,y)=\frac{\partial H(x,y)}{\partial x}$ with respect to $x$ are relatively prime, provided that $H'(x,y)\ne0$. The Bezout Lemma then yields polynomials $A(x,y)$, $B(x,y)$ and a nonzero polynomial $c(y)$ such that $A(x,y)H(x,y)+B(x,y)H'(x,y)=c(y)$. So whenever $c(a)\ne0$, then $H(x,a)$ is still separable, so in particular it is not a square.
Remark: As pointed out by Will Sawin, the argument needs to be enhanced if $K$ has positive characteristic. Since I don't know whether the OP is interested in that too, and because the treatment becomes somewhat more technical, I'll leave that case open.
