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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!

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    $\begingroup$ 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 $\epsilon<1$. $\endgroup$ – Philipp Lampe Mar 14 at 14:22
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    $\begingroup$ The lemma is in fact true. Note that you can take a large constant C > 0 in this case. $\endgroup$ – Marcelo Nogueira Mar 14 at 14:31
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    $\begingroup$ @MarceloNogueira: for what value of $C_{1/2}$ is $C_{1/2}K^{1/2}>K+1$ for all $K$? $\endgroup$ – Alex B. Mar 14 at 17:38
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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 $K<k\leq 2 K$, $L<l\leq 2 L$. (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

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    $\begingroup$ Nell, thanks so much for your explanation ! $\endgroup$ – Marcelo Nogueira Mar 14 at 18:27

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