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I suppose a conjecture implies this so there might be an unconditional proof.

Let $F(x,y)=0$ be a curve with infinitely many integral points $(u^m,v^n)$ where $\gcd(u,v)=1$ infinitely often and $m \ge 3,n \ge 2$. Such curves are easy to construct by starting with a parametrization for example.

For a bivariate polynomial $F$ define $\operatorname{High}(F)$ to be the sum of the highest degree monomials (i.e., $\operatorname{High}(F) = F$ iff $F$ is homogeneous and $\operatorname{High}(2x^3+3y^3+x+y)=2x^3+3y^3$). Let $\gcd(\operatorname{High}(F),xy)=1$.

Under these conditions is $\operatorname{High}(F)$ not square-free?


This can't be relaxed to $m \ge 2 $

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(Added some remarks from my comments)

Suppose that $F=F_1F_2\cdots F_r$ with $F_i$ irreducible. If $F=0$ has infinitely many points of the requested form, then so does one of the factors. Furthermore, $\text{High}(F)=\text{High}(F_1) \text{High}(F_2)\cdots\text{High}(F_r)$. So in order to look at the question, we may assume that $F$ is irreducible. But the $F$ is even absolutely irreducible, which we assume from now on.

As $F(x,y)=0$ has infinitely many integral solutions, then by Siegel's Theorem the projective closure of this curve has at most $2$ points at infinity. These point are just those with coordinates $(x:y:0)$ with $H(x,y)=0$, where $H=\text{High}(F)$. So the polynomial $H$ has total degree at most $2$ if it is to be separable and not divisible by $x$ nor $y$.

So a counterexample (that is where $H(x,y)$ is not squarefree) would have degree at most $2$.

I believe that degree $2$ can be ruled out. However, degree $1$ seems to amount to solve Pillai's conjecture: Given nonzero integers $A,B,C$, then $Au^m+Bv^n=C$ has only finitely many integral solutions $u,v,m,n$ with $m,n\ge3$. You have the additional assumption that $u$ and $v$ are relatively prime. I'm sure that this doesn't make the conjecture easier.

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You don't use the hypothesis about perfect powers and in particular $m \ge 3$? You don't need this hypothesis to answer the question? – joro Aug 20 '13 at 6:32
You don't mention coprimality at all. You don't need it to answer the question? – joro Aug 20 '13 at 6:59
@joro: Indeed, it doesn't seem to be necessary to use these additional assumptions in order to reduce to degree $2$ which is easy to handle. – Peter Mueller Aug 20 '13 at 7:57
Hm, isn't $x^n-y^n=0$ counterexample to your claim about degree 2?. Points are (t,t). High(x^n-y^n)=x^n-y^n, the degree is n and it is squarefree? – joro Aug 20 '13 at 8:21
Well, but this curve contains a straight line. Trivial examples like this are always possible. However, as your function $\text{High}$ is multiplicative, your question only makes sense for irreducible polynomials. – Peter Mueller Aug 20 '13 at 9:13

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