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Let $x$ be a sufficiently large positive number. Do there exist positive integers $d_1, d_2$ with the following properties?

  1. $d_1, d_2$ are both $12 \log^2x$-smooth, i.e., all prime divisors of $d_1, d_2$ are at most $12 \log^2 x$ in size;

  2. $d_1, d_2 \asymp x$;

  3. For all integers $1 \leq k \leq x$, either $k$ is a perfect square or $k$ is not a square residue to both $d_1$ and $d_2$.

  4. $\gcd(d_1, d_2) = 1$.

Stanley Yao Xiao
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    This is not even close to being possible, even if we only consider 2 and 3 -- there are $\gtrsim \frac{x}{2^{\omega(d_1)}}\gtrsim x^{1-o(1)}$ square residues mod $d_1$ up to $x$ and only $\sim\sqrt{x}$ squares. I would imagine it is impossible even if we require $k$ to be a non-square to at least one of $d_1$ and $d_2$. – Aleksei Kulikov Mar 03 '24 at 06:13
  • why not write condition 2 as $d_1,d_2 \asymp \sqrt{x}$? – mathworker21 Mar 03 '24 at 07:34
  • Sorry I mis-calibrated, I want both $d_1, d_2$ to be size $x$, not $\sqrt{x}$. – Stanley Yao Xiao Mar 03 '24 at 07:57
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    Still, almost the same logic applies, we just need some bound on the number of non-residues in intervals, which can be extracted from the bounds on character sums. Say, Polya--Vinogradov, should give that we need $d_1 \ge x^{2-o(1)}$ even if we ignore $d_2$ completely, and the Burgers bound might give even bigger lower bound (I don't know what is the truth, it might be close to exponential). – Aleksei Kulikov Mar 03 '24 at 09:09
  • @AlekseiKulikov could you write out the argument in detail? A priori it does not seem impossible, since there are $O(x 2^{-\omega(d_1)})$ square residues modulo $d_1$ and almost all integers in $[1, x]$ are non-residues, so it does not seem impossible that all of the residues mod $d_1$ sans squares are contained in the non-residues modulo $d_2$. – Stanley Yao Xiao Mar 03 '24 at 09:32

1 Answers1

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We will ignore most of your assumptions, let $d = d_1$ be a number and assume that for $1 \le k \le x$ we have that $k$ is a square modulo $d$ if and only if $k$ is a square in $\mathbb{N}$. We will show that $d$ is at least $x^{2-o(1)}$ (with Burgess bound we should be able to improve $2$ to something slightly bigger, probably up to $4$).

We will assume that $d$ is smaller than that and simply show that the number of quadratic residues modulo $d$ up to $x$ is at least $x^{1-o(1)}$, which is bigger than the number of squares (which is at most $\sqrt{x}+1$).

Let us for simplicity assume that $d$ is squarefree (there are some annoying but not insurmountable technical issues otherwise). Consider all quadratic characters $\chi_1, \ldots, \chi_m$ modulo $d$, where $m = 2^{\omega(d)}$ and $\chi_1$ is a trivial character: $\chi_1(k) \equiv 1$ if $(k, d) = 1$ and $0$ otherwise. Consider the sum $$f(k) = \chi_1(k) + \ldots + \chi_m(k).$$

I claim that if $k$ is not a square modulo $d$ then this sum is $0$. Indeed, this sum can be expressed as a product over primes $p$ dividing $d$ of $(1+\chi_p(k))$ and $\chi_p(k) = -1$ precisely if $k$ is a non-residue modulo $p$. So, we have $f(k) = 0$ if $k$ is a non-residue modulo $d$ and $f(k) \le 2^{\omega(d)}$ in general. Thus, the number of residues from $1$ to $x$ is at least $$\frac{1}{2^{\omega(d)}}\sum_{k = 1}^x f(k) = \frac{1}{2^{\omega(d)}}\sum_{s=1}^m \sum_{k=1}^x \chi_s(k).$$

The sum for $s = 1$ gives us the number of $1 \le k \le x$ which are coprime to $d$, which is $\frac{x\phi(d)}{d} + O(2^{\omega(d)})$, see, e.g., here distribution of coprime integers . All other sums are $O(d^{1/2 + o(1)})$ by Polya--Vinogradov, so sum of sums is $\frac{x\phi(d)}{d2^{\omega(d)}} + O(d^{1/2+o(1)})$. If $d \le x^{2-o(1)}$ then the main term is bigger than twice the error term and so the sum is $x^{1-o(1)}$.

For non-squarefree $d$ we have to first of all cover the non-trivial characters modulo powers of $2$, and then the simplest approach for powers of $p$ is to just consider any number divisible by $p$ a non-residue (then we will have the sum of $3^{\omega(d)}$ terms, and the main term would be slightly smaller, but only by $d^{o(1)}$, so still miles bigger than the error term).

In fact, some google-fu led me to a blog post by Terence Tao https://terrytao.wordpress.com/2013/06/22/bounding-short-exponential-sums-on-smooth-moduli-via-weyl-differencing/ from which it appears that for smooth modules the bound for the character sums can be further improved, so likely this can give an even bigger lower bound for $d$ if we take your first point into account.

Aleksei Kulikov
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