Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

Counterexamples?

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

share|improve this question

1 Answer 1

(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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.