Does an increasing number of solutions modulo p imply infinite integer solutions?

Question

I'm working on a problem involving modular arithmetic, which I've never touched before to be fair, that's why I need help.

Consider an equation with two variables, let's call them x and y, in modular arithmetic:

f(x) - g(x, y) ≡ 0 (mod p)

where p is a prime number and x,y $\in ℤ$.

Now, suppose I can prove the following:

1. For every prime p, there exists at least one solution to this equation.
2. As p increases, the number of solutions to this equation also increases.

My question is: Can I argue that if the number of solutions is always positive and increasing with p, this implies that there are infinitely many solutions to the equation in the set of integers (not modulo any prime)?

In other words, does the behavior in modular arithmetic give me grounds to claim the existence of infinitely many integer solutions?

So I'm particularly interested in this Questions:

1. Is this a valid line of reasoning?
2. If not, what additional conditions or proof elements would be needed to make such a claim?
3. Are there any known theorems or results that relate the behavior of solutions in modular arithmetic to the existence of integer solutions?

I'm reaching out for help as I'm not studying mathematics and my background in this area is quite limited.
Thanks for any help

Answer 1

The number of solutions of $x^8-16y^8\equiv0\bmod p$ goes to infinity with $p$, but $x^8-16y^8=0$ has no integer solution other than $x=y=0$.

See Showing $x^8\equiv 16 \pmod{p}$ is solvable for all primes $p$ for a proof that $u^8\equiv16\bmod p$ has a solution $u$ for every prime $p$. Given any prime $p$, and any integer $y$, let $x=uy$, then $x^8\equiv16y^8\bmod p$.

Answer 2

I am assuming $f$ and $g$ are polynomials. The question touches some quite deep number theory, arithmetic geometry in particular:

• Faltings's theorem basically says that "generically" (the precise condition is about a geometric quantity called the genus) if both $f$ and $g$ are high enough degree polynomials (in the absence of singularities on the projective version degree four suffices) and $f-g$ does not factor, then there can be only finitely many solutions with $x,y\in\Bbb{Q}$. As Aphelli explained in a comment, the earlier Siegel's theorem on integral solutions suffices for the current discussion.
• On the other hand, Weil conjectures proved by Deligne (with significant contributions from others, IIRC Weil himself had sufficient results to cover the case of two variable equations) tell that, again generically, the number of pairwise non-congruent solutions $(x,y)$ modulo a prime $p$ differs from $p$ by at most a constant times $\sqrt p$.

It may be possible to get around these obstacles in some special cases (at least when $f-g$ factors), but they show that you should expect the exact opposite to be true: the number of modular solutions always tends to infinity as $p$ grows, but there are only finitely many solutions in $\Bbb{Z}$.

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Gerry already gave an interesting example. I think the equation $$x^2+y^2+1=0$$ also works. It obviously has no solutions in $\Bbb{R}$ or $\Bbb{Z}$, but the number of solutions modulo a prime $p>2$ is $p\pm 1$. The Hasse principle described in a comment by Billy largely holds for quadratic equations.

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It is not unnatural to think that "lifting" a modular solution to one in $\Bbb{Z}$ could be possible. After all, if $p_1<p_2<\cdots<p_n$ are primes and $(x_1,y_1)$, $(x_2,y_2)$, $\ldots$, $(x_n,y_n)$ are solutions modulo those primes, then the Chinese remainder theorem promises the existence of integers $x,y$ such that $x\equiv x_i$ and $y\equiv y_i$ modulo $p_i$, when $(x,y)$ is automatically a solution modulo the product $p_1p_2\cdots p_n$.

Furthermore, by a process called Hensel lifting it is often possible to also find solutions modulo powers of the primes, ${p_i}^{a_i}$, before calling upon the Chinese remainder theorem. Anyway, what typically happens is that by adding more primes to the mix, the sequence of solutions CRT produces won't converge to an integral solution. In view of the deep results above, this is to be expected.