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Suppose I have a mystery number $m$ modulo $p$ that I wish to find. I know the value of $m+x_i^2$ where $x_i$ is randomly chosen modulo $p$ for some large number of different $x_i$, $N$ many, $N \gg \log(p)$.

Is it possible to deduce the value of $m$?

If we replace $m+x_i^2$ with $m+C_i*x_i^2$ for known $C_i$, can we find $m$?

I believe the answer should be yes in both cases, there are roughly $p/2$ quadratic residues so each question gives around $1$ bit of information. There are lattice and Fourier based solutions on the linear version of the problem with $m+K_ix_i$ where, lets say the first bit of $x$ is known modulo $p$ with probability $\varepsilon > 0$. (Bleichenbacher’s algorithm).

https://eprint.iacr.org/2013/346.pdf

I was thinking you can use Gauss sums or something to tease out the value of $m$ slowly from $x^2+m$. Something like computing an average of $e(m+x^2)$ where $e(x)$ is the complex exponential and having it converge to $m$. But I don't know enough about Gauss sums.

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  • $\begingroup$ Proper notation is $N\gg\log(p),$ not $N>>\log(p).$ I edited accordingly (and also took the liberty of changing $\epsilon$ to $\varepsilon$). $\endgroup$ Commented Oct 21, 2023 at 21:14
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    $\begingroup$ What's known here? Is $p$ known? are the $x_i$ known? $\endgroup$ Commented Oct 22, 2023 at 5:38
  • $\begingroup$ $p$ is known, $x_i$ are unknown, $m$ is unknown, but the values $m+x_i^2$ are given to you. It's very surprising that its possible to find $m$ even in the linear case. $\endgroup$ Commented Oct 22, 2023 at 12:26
  • $\begingroup$ If $Q$ is the set of quadratic residues, then $|Q\cap (Q+a)|$ is roughly $p/2$, which shows that roughly $p/2$ values suffice. $\endgroup$ Commented Oct 23, 2023 at 11:35
  • $\begingroup$ Yes, but how would one go about approximating the intersection set, actually finding the intersection? $\endgroup$ Commented Oct 23, 2023 at 16:02

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