Let $F(x,y)$ be a squarefree binary form with integer coefficients, possibly reducible, $\deg(F) \ge 3$.

I am interested in ways of getting infinitely many integer solutions $(x,y,m), m \ne 0$ to $F(x,y)=m$, maximizing $\max(|x|,|y|)$ relative to $m$.

More formally, suppose $F$ is as above, $f$ is an increasing function, and for infinitely many $x,y,m$, $m \ne 0$ the following hold:

1) $F(x,y)=m$.

2) $\max(|x|,|y|) \ge f(|m|,\deg(F))$ or $\max(|x|,|y|) \ge f(|m|)$

How fast can $f$ grow?

Partial result: For $\deg(F)=3$, it is possible to have $f(m)=C m$ for $C$ an arbitrary large constant depending on $F$. This may be best possible for $F$ of degree $3$.

Added later The construction with $f(|m|)= C m$ as suggested in comments:

For $C \sim \varphi^{k+1}$, we can take $f(|m|)= C m$ and $$G(x,y) = (F_{k+1} x - F_k y) (x^2 + x y - y^2)$$ where $F_k$ is the $k^{th}$ Fibonacci number.

Then we can take $x=F_n$, $y=F_{n+1}$, $m=(-1)^{n+1} F_{n-k}$.

We have the identities

$$ F_{k+1} F_n - F_k F_{n+1} \ = \ F_{k+1} x - F_k y \ = \ F_{n-k} $$ $$ F_n^2 + F_n F_{n+1} - F_{n+1}^2 \ = \ x^2 + xy - y^2 \ = \ (-1)^{n+1}.$$

which yield $G(x,y) = m$.

This construction seems to give an unbounded number $G(x_i,y_i)$ of unbounded quality for the Granville-Langevin conjecture, which is equivalent to $abc$ without using the radical at all.

This is a lower bound for $f(|m|)$ and experimentally there are much better solutions with the same $G$, though I don't know if they are infinite (probably not).

Multiplying $G$ by similar linear factors, I believe one can get $f(|m|)=C |m|^{\frac{1}{(\deg(G)-2)}}$.

  • $\begingroup$ In 1), do you mean "There are infinitely many $m$ such that $F(x,y)=m$ has integer solutions"? In 2), do you intend to take the maximum over all solutions, not just over a pair which constitutes one solution? $\endgroup$ – GNiklasch May 12 '14 at 12:07
  • $\begingroup$ @GNiklasch yes, exactly. As I wrote $x,y,m$ are integers. $\endgroup$ – joro May 12 '14 at 12:08
  • $\begingroup$ My point was to clarify what you're quantifying over and how. "Thue equation" connotes that one nonzero value of $m$ is picked and kept fixed. So you're really looking at infinitely many Thue equations built around the same binary homogeneous form, right? $\endgroup$ – GNiklasch May 12 '14 at 12:14
  • $\begingroup$ @GNiklasch yes, $F$ is fixed and $m$ varies, so these indeed are infinitely many Thue equations.In addition I allow reducible $F$. $\endgroup$ – joro May 12 '14 at 12:26
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    $\begingroup$ @GerryMyerson edited the question with unbounded C. $\endgroup$ – joro May 13 '14 at 12:27

If the degree of the form is $d$, I think you get solutions with $f(|m|) > cm^{1/(d-2)}$ by applying Dirichlet's theorem in diophantine approximation to a root of $F(x,1)=0$. And you cannot do better than $cm^{1/(d-2)+\epsilon}$ by Thue-Siegel-Roth.

  • $\begingroup$ Thank you. Would you please explain in more detail why one can't do better? $\endgroup$ – joro May 17 '14 at 8:35
  • $\begingroup$ $F(x,y)=m$ is the same as $F(x/y,1)=m/y^d$ and, since $F$ is squarefree, $F(x/y,1)$ is, up to a constant, close to $(x/y - \alpha)$ for some root $\alpha$ of $F(x,1)=0$. Now apply Thue-Siegel-Roth. $\endgroup$ – Felipe Voloch May 17 '14 at 12:40
  • $\begingroup$ Thank you. For the construction do you mean real root? Little can be done about $F(x,y)=x^4+y^4$. $\endgroup$ – joro May 18 '14 at 13:11
  • $\begingroup$ You are correct, the construction only applies if there is a real, irrational, root. $\endgroup$ – Felipe Voloch May 18 '14 at 14:03

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