Upper bound for the number of solutions of a Diophantine equation

Consider the Diophantine equation

$$k^2 + k - \sigma (\ell^2 + \ell) = m,$$ where $$N \leq k \leq 2 N$$, $$L \leq \ell \leq 2 L$$, $$m \in \mathbb{Z}$$ and $$\sigma \in \mathbb{R}$$.

For which values of the parameter $$\sigma$$ is it possible to say that the number of solutions of this equation is bounded by $$O(\min(N, L)^{\epsilon})$$ ?

To facilitate progress, I'm going to put this auxiliar lemma.

$$\underline {Lemma}$$. For every $$\varepsilon > 0$$, there exists $$C_{\varepsilon} > 0$$ such that, for every $$m \in \mathbb{Z}$$ and $$K$$ positive integer,

$$\sharp \{(x, y) \in \mathbb{N}^{2} \mid K \leq x \leq 2 K , x^{2} \pm y^{2} = m \} \leq C_{\varepsilon} K^{\varepsilon}.$$

I would be very grateful for any suggestion!

• The lemma is false. Suppose that $m=0$. Then every pair $(x,x)$ with $K\leq x\leq 2K$ belongs to the set on the LHS. Hence, this set contains $K+1$ elements and so its cardinality cannot be bounded from above by $C_{\epsilon} K^{\epsilon}$ for $\epsilon<1$. – Philipp Lampe Mar 14 at 14:22
• The lemma is in fact true. Note that you can take a large constant C > 0 in this case. – Marcelo Nogueira Mar 14 at 14:31
• @MarceloNogueira: for what value of $C_{1/2}$ is $C_{1/2}K^{1/2}>K+1$ for all $K$? – Alex B. Mar 14 at 17:38

First of all, you can assume $$\sigma\in \mathbb{Q}$$. Otherwise, the only solutions there can be are those with $$l^2+l=0$$ and $$k^2+k=m$$, and there's certainly only a bounded number of those.

Multiplying by the denominator, we see that we are being asked for a bound on the number of solutions to $$a (k^2 + k) + b (l^2 + l) = c$$ with $$k$$, $$l$$ in dyadic intervals. We multiply both sides by $$4$$ and add $$a+b$$ to complete the squares. Thus we reduce our problem to that of bounding the number of solutions to $$a k^2 + b l^2 = c$$ with $$k$$, $$l$$ in dyadic intervals. Multiplying by $$a$$ and then replacing $$a k$$ by $$k$$, we reduce our problem to that of bounding the number of solutions to $$k^2 + d l^2 = n$$ for given $$d$$ and $$n$$, with $$k$$ and $$l$$ in dyadic intervals $$K, $$L. (I take the implied constant in the bound you wish is allowed to depend on $$\sigma$$. The bound I will give will depend on $$d$$, though not on $$n$$.)

For $$d>0$$, the number of solutions is obviously bounded by the number of ideals of norm $$n$$ in the ring of integers of $$\mathbb{Q}(\sqrt{d})$$. That number is bounded by the number of divisors of $$n$$, which is $$O_\epsilon(n^\epsilon) = O_\epsilon(\max(K,L)^\epsilon)$$. To obtain the bound $$O_\epsilon(\min(K,L)^\epsilon)$$, note that, if $$L^2 < 2K/3d$$, there can be at most one solutions to your equation, as two consecutive values of $$k^2$$ differ by at least $$2 K + 1$$, and $$d (2 L)^2 - d L^2 = 3 d L^2$$.

For $$d<0$$, you also have to take quadratic units in $$\mathbb{Q}(\sqrt{d})$$, but, as there is only a logarithmic number of them in a box of given size (the group of units being isomorphic to $$\mathbb{Z}$$ times bounded torsion), you still get a bound of $$O_\epsilon(\max(K,L)^\epsilon)$$, and hence of $$O_\epsilon(\min(K,L)^\epsilon)$$.

The only exception is given by $$n=0$$ and $$d$$ of the form $$-r^2$$, $$r$$ an integer. Then the number of solutions to $$k^2+d l^2 = n$$ is evidently infinite, and the number of solutions in a box is linear on the size of the box. It is easy to see that this is the case of $$c=-(a+ b)/4$$, $$a b=-r^2$$ in the equation $$a (k^2+k) + b (l^2+ l) = c$$. That case corresponds to $$\sigma\in \mathbb{Q}^2$$, $$m = (\sigma-1)/4$$ (and thus $$\sigma\in \mathbb{Z}^2$$) in the original problem.

tl;dr a simple exercise in quadratic number fields

• Nell, thanks so much for your explanation ! – Marcelo Nogueira Mar 14 at 18:27