# Large solutions to Thue equations

Suppose that $f(x,y) \in \mathbb{Z}[x,y]$ is a homogeneous polynomial, or binary form, of degree $d$. The equation $$f(x,y) = h$$ for a given integer $h$ is known as Thue's equation (so named because Thue showed that this equation always has a finite number of solutions in integers $x,y$). There have been much work done subsequently to improve the bound on the number of solutions. As far as I know, the best general result is due to Stewart, who showed that the number of solutions is no more than $$2800\left(1 + \frac{1}{8 \varepsilon d}\right)d^{1 + \omega(g)},$$ where $\varepsilon$ is some positive constant and $g$ is a divisor of $h$ that is sufficiently large with respect to the discriminant of $f$ (and other invariants). One should interpret the above result as saying that when $h$ has a very large prime divisor or a large prime power divisor, the result is essentially that the number of solutions is $O(d^2)$. Indeed, in cases when $g$ can be chosen as 1, we even get the bound $O(d)$, which we know is best possible in general (due to an example of Bombieri and Schmidt, for example).

I am particularly interested in the case when $h$ is small, but the solution $(x_0, y_0)$ is large. It is not too hard to see that one of $x_0$ or $y_0$ can be made arbitrarily large in some cases. For instance, consider the equation $$(m^{kd} + 1)x^d - y^d = 1$$ which obviously has the solution $(1, m^k)$. Hence there are equations where $y$ can be arbitrarily large compared to $h$.

However, note that while $y$ is large with respect to 1, it is small compared to the coefficients. My question concerns this type of behaviour.

1) Suppose that a Thue equation $f(x,y) = h$ has a solution with $y$ (say) being much larger than $h^{1/d}$. Do we necessarily have that the coefficients of $f$ have to be comparably large?

2) Suppose we have a Thue equation $f(x,y) = h$ where $h$ is very small compared to the height $\lVert f \rVert$ of $f$, which is the largest of the absolute values of the coefficients of $f$. Is there a uniform bound (that is, independent of even the degree of $f$) that can be established for the number of solutions where at least one coordinate is very large compared to $h$ (but not necessarily to $\lVert f \rVert$)?

Thank you for any insights.

• How do you define "much larger" for h^(1/d)? – joro Dec 6 '14 at 10:18

I'm not sure whether the second question is inadvertently misworded. As it is, the answer can easily be shown to be no. Fix $h$ and let $A$ be huge compared to $h$. Let $$f(x,y)=A(x-y)(x-2y)\cdots (x-ny)+h y^n.$$ Then the Thue equation $f(x,y)=h$ has the $n$ solutions $(1,1),(2,1),\dots,(n,1)$. Here $h$ is very small compared to $\lVert f \rVert \gtrsim n! A$. So we cannot expect a bound on the number of solutions independent of the degree of $f$.

Added in response to the edited version of Question 2: The answer is still no. Take $n$ to be huge compared to $h^2$. Now in the above, the solutions $(\lfloor \sqrt{n} \rfloor,1),\dots,(n,1)$ all have a coordinate that is huge compared to $h$ and their number is roughly $n$.

• I added a condition to eliminate this particular counter-example, which as you noted is quite obvious. – Stanley Yao Xiao Dec 7 '14 at 1:21

I think you need $d>2$.

The answer to the first question is "no". For $F_n$ the Fibonacci numbers, $x,y=F_{2n},F_{2n+1}$ satisfy $x^2+xy-y^2+1=0$

Take the degree $3$ Thue equation $f(x,y)=x(x^2+xy-y^2)$.

For Fibonacci numbers as above take we have $f(x,y)=x$.

Take $h=x$. We have $x \gg x^{1/3}$ and all coefficients are units.

If you want $f$ irreducible, take $d=3$ and $f$ with real irrational root, then use Voloch's answer here

To the edited question about uniform bound.

I think again no.

In Siksek's answer replace $x-my$ by $x-m^{2^k}y$. You have the solutions $(m^{2^k},1)$ for fixed $h$.