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The number of integer points with a given norm in the integer grid $\mathbb{Z} \times \mathbb{Z}$ may be calculated via the generating function $$\theta_3(q)^2= \left(\sum_{n \in \mathbb{Z}} q^{n^2}\right)^2 = 1+4 q + 4 q^2 + 4 q^4 + 8 q^5 + \cdots,$$ which is a so called (squared) Jacobi function. Similarly the generating function for the shifted $(\mathbb{Z}+a)\times(\mathbb{Z}+b)$ is given by (say $\Theta(a,b)$) $$\Theta(a,b)=\left(\sum_{n \in \mathbb{Z}} q^{(n+a)^2}\right)\left(\sum_{n \in \mathbb{Z}} q^{(n+b)^2}\right)$$ which can also be written as a function as Jacobi function of two variables.

For example, if $(a,b)=(\sqrt{2}/3,3/10)$, we have

$$\Theta(a,b) \approx q^{1.91222}+q^{0.769413}+q^{0.712222}+q^{0.369413}+q^{0.312222}$$

My question is if the following problem (or anything related) has been studied: what is the value of (a,b) that makes (i) all coefficients in the expansion of $\Theta(a,b)$ equal to one and (ii) the difference between successive exponents in $\Theta(a,b)$ the largest possible.

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  • $\begingroup$ What is the "difference between successive exponents"? Which ones? $\endgroup$
    – Igor Rivin
    Aug 1, 2017 at 23:27

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Won't the following argument show that the difference between successive exponents can never be bounded away from zero no matter how clever you try to be in selecting $(a,b)$?

The idea is to consider pairs of integers $(n,m)$ such that $na-mb$ is close to zero.

  1. If the ratio $a/b$ is rational then we can find integers $n,m$ such that $na=mb$. But in that case $$(n+a)^2+(-m+b)^2=n^2+m^2+a^2+b^2+2(na-mb)=n^2+m^2+a^2+b^2$$ and also $$ (-n+a)^2+(m+b)^2=\cdots=n^2+m^2+a^2+b^2 $$ meaning that the theta series has coefficients $>1$.
  2. On the other hand if the ratio $a/b$ is irrational then, to a given $\epsilon>0$, we can find integers $n,m$ such that $$ |na-mb|<\epsilon. $$ This is because the additive group generated by $a$ and $b$ is then a dense subset of $\Bbb{R}$. But, reusing the above points, we see that $$ \begin{aligned} ||(n+a,-m+b)||^2-||(-n+a,m+b)||^2&=(n^2+m^2+a^2+b^2)+2(na-mb)\\ &-(n^2+m^2+a^2+b^2)+2(na-mb)\\ &=4(na-mb), \end{aligned} $$ which is $<4\epsilon$.

A geometrical motivation for finding these points came from the observation that when $\pm(n,-m)$ is nearly orthogonal to $(a,b)$, we are bound to get two points yielding nearly equal exponents of $q$.

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  • $\begingroup$ Very true, I think I was overly optimistic to find a lower bound for the ``infinite'' lattice. The problem would only make sense for a finite (e.g. inside a cube) subset of $\mathbb{Z}^2$. $\endgroup$
    – Campello
    Aug 3, 2017 at 14:35
  • $\begingroup$ @Campello. Unfortunately that is the case. It looks like you will have to contend with optimizing how quickly the differences between the exponents go to zero as the range of $n,m$ increases. Not entirely unlike in my past question. $\endgroup$ Aug 3, 2017 at 14:42
  • $\begingroup$ Yes, my question was clearly not well formulated... :P Thanks for the answer ! $\endgroup$
    – Campello
    Aug 3, 2017 at 22:24

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