Suppose that we have a binary form $f(x,y) \in \mathbb{Z}[x,y]$ of degree 4, and that we explicitly have

$$\displaystyle f(x,y) = a_0 x^4 + a_1 x^3 y + a_2 x^2 y^2 + a_3 xy^3 + y^4,$$ so that $(0,1)$ is a solution to the equation $f(x,y) = 1$ (most notably, we require that the equation $f(x,y) = 1$ have a solution in $\mathbb{Z}^2$).

Suppose we define the auxiliary form $g = u_0 x^4 + \cdots + u_4 y^4$ by $u_0 = a_0$,

$u_1 = a_1 - 4 a_0 \left( \frac{b-c}{p^k}\right)$,

$u_2 = a_2 - 3a_1\left(\frac{b-c}{p^k}\right) + 6a_0 \left(\frac{b-c}{p^k}\right)^2$,

$u_3 = a_3 - 2a_2 \left(\frac{b-c}{p^k}\right) + 3a_1 \left(\frac{b-c}{p^k}\right)^2 - 4a_0 \left(\frac{b-c}{p^k}\right)^3$ and $u_4 = a_4 - a_3 \left(\frac{b-c}{p^k}\right) + a_2 \left(\frac{b-c}{p^k}\right)^2 - a_1 \left(\frac{b-c}{p^k}\right)^3 + a_0 \left(\frac{b-c}{p^k}\right)^4 $

where $p$ is a prime, $k \geq 1$ is an integer, and $b,c$ are such that each $u_i$ is an integer and $\gcd(u_0, \cdots, u_4) = 1$. Can one bound the number of choices of $b,c$ by an absolute constant for which $g(x,y) = 1$ also has a solution in $\mathbb{Z}^2$?

My motivation is the following situation. Stewart (my PHD advisor) conjectured in 1991 in "On the number of solutions of polynomial congruences and Thue equations", Journal of the American Mathematical Society, (4) Volume 4 (1991), 793-835 that there exists a constant $c_0$ such that for a given binary form $f(x,y) \in \mathbb{Z}[x,y]$ of degree $d \geq 3$, there exists a number $r(f) \in \mathbb{N}$ such that for all integers $h$ with $|h| \geq r(f)$, the number of primitive solutions to the equation $f(x,y) = h$ is bounded by $c_0$. A weaker form of the conjecture is to replace the number $c_0$ (which does not depend on the degree $d$) with a quantity $c(d)$ which depends on $d$. In the same paper, Stewart counted the solutions to $f(x,y) = h$ by considering a form $f(p^k x + by, y)$ with $p^k || h$ and where $b \pmod{p^k}$ is chosen so that the form $\tilde{f}(x,y) = f(p^k x + by, y)$ has content $p^k$. Then one obtains an equation of the form $f'(x,y) = h/p^k$, and continuing in this manner one eventually gets to a counting problem that one can solve. The above construction in my question 'reverses' this process, by first starting with a form $f(x,y)$ which manifestly has a solution to the equation $f(x,y) = 1$, then trying to lift the solution to one of the form $f'(x,y) = p^k$. My question is basically asking under what circumstances and two distinct forms (in my case, both degree 4) $f,g$ lift to the same form $f'$.

Edit 06 Sep 2014: Fixed the expressions for $u_i$, made calculation errors in the original.

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    $\begingroup$ Can you say something about why this form $g$ might be a natural thing to consider? Also, unless I've misread the formulae, replacing $(b, c)$ with $(b + \lambda, c + \lambda)$ for any integer $\lambda$ does not change the form $g$, so the number of valid pairs $(b, c)$ is either 0 or infinite. $\endgroup$ Commented Sep 1, 2014 at 23:32
  • $\begingroup$ I added my motivation to the question in the body of the question. $\endgroup$ Commented Sep 1, 2014 at 23:59

1 Answer 1


Have you left some condition off of Stewart's conjecture? It's not right as you've stated it. Indeed:

Theorem Let $f(x,y)\in\mathbb{Z}[x,y]$ be a polynomial of degree $3$ with $\operatorname{Disc}(f)\ne0$. Then for every $N$ there exists a non-zero integer $m$ such that the equation $f(x,y)=m$ has at least $N$ solutions $(x,y)\in\mathbb{Z}^2$.

If I recall correctly, this was noted by Chowla. The idea is to choose an $m_0$ so that $f(x,y)=m_0$ has infinitely many rational solutions, which is easy to do since it's a genus 1 curve, so just choose $m_0$ so the rank is positive. Take $N$ rational points $(x_i,y_i)=(a_i/d_i,b_i/d_i)$. Then take $m=m_0(d_1d_2\cdots d_N)^3$ and the curve $f=m$ will have $N$ integer points.

Possible corrections. If you only take solutions with $\gcd(x,y)=1$, or if you require that $m$ be $d$'th power free (where $d=\deg(f)$), Cam's conjecture might be true. Ditto for $\deg(f)\ge4$, although then Caporaso-Harris-Mazur predicts (assuming Bombieri-Lang) that there's a uniform bound for the number of rational points (if you fix $\deg(f)$).

Finally, I'll mention that if we let $C_m$ be the curve $f(x,y)=m$ and $J_m$ its Jacobian, then it is known that the number of integer solutions with $\gcd(x,y)=1$ is bounded by $C_f^{1+\operatorname{rank} J_m(\mathbb{Q})}$, where the constant $C_f$ is independent of $m$. Of course, that's potentially weaker than Cam's conjecture, although some people believe that ranks in twist families are bounded.

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    $\begingroup$ Cam assumed that $\gcd (x,y)=1$. $\endgroup$ Commented Sep 2, 2014 at 3:46
  • $\begingroup$ Yes, I forgot to put in the primitive condition in there, my apologies $\endgroup$ Commented Sep 2, 2014 at 11:46

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