Let $f\in\Bbb Z[x_1,...,x_n]$ be a polynomial with no integer solutions. We call such $f$ not solvable $\pmod p$ if for some prime number $p$ the polynomial $f$ does not have a solution $\pmod p$. This question seems extremely broad, but I am looking for a explict description for the set of such $f$. This property of $f$ is a commonly used way to (possibly) deduce a diophantine equation has no solutions. The simplest example is $3x+3y=2$, when looking at solutions $\pmod 3$.

Another not-so-trivial example is given by $7x^3+2=y^3$, by looking at the equation $\pmod 7$ one has $y^3=2\pmod 7$ but then $1=y^6=4\pmod 7$ a contradiction, by Fermat's little theorem, since appearently $y\neq 0\pmod 7$. In general there are only finite cases to examine, and often times multiplicative structure of $\Bbb Z/p\Bbb Z$ helps us reduce the amount of calculations needed.

Observe that a polynomial $f$ not solvable $\pmod n$ for some integer $n>1$ then by Chinese Reminder Theorem there is a prime power for which $f$ is not solvable $\pmod {p^n}$ and if Hensel's Lemma applies(that is, $f'$ with no repeated roots modulo $p$) $f$ is not solvable $\pmod p$ for some prime $p$. So that the definition we made in the first paragraph is almost general enough.

A weaker description/partial result related to this property are also appreciated, since by my guess this should have been extensively studied centries ago. Thank you very much.

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    $\begingroup$ Also in your 3rd paragraph what you state is incorrect, consider $x^2 + 1$ mod $4$ and mod $2$... $\endgroup$ – Mummy the turkey Oct 6 '20 at 22:21
  • $\begingroup$ @Mummytheturkey Sorry for the handwaving I should have said that $f(x)$ with $f'(x)$ not $0$ modulo $p$. $\endgroup$ – William Sun Oct 7 '20 at 0:02

For polynomials in one variable this is, as far as I know, a wide open question. If $f(x) \in \mathbb{Z}[x]$ has the property that its splitting field is abelian then by the Kronecker-Weber theorem it embeds in some cyclotomic field and the splitting behavior of $f(x) \bmod p$ is governed by congruence conditions on $p$. In particular if $f(x) = x^2 - q$ this question is answered by quadratic reciprocity which more or less completely answers this question if $f$ is quadratic. This is a special case of class field theory, which generalizes quadratic reciprocity to Artin reciprocity.

If $f$ is cubic, irreducible, and has Galois group $S_3$ (the smallest nonabelian group) then the situation is already enormously more complicated. In this case the splitting behavior of $f(x) \bmod p$ is described (conjecturally? I don't know what the state of the art is, and we might also need $f$ to have a complex root) by the coefficients of a modular form, as described e.g. here. As an explicit example, if $f(x) = x^3 - x - 1$ then $f(x)$ has a root $\bmod p$ if and only if the coefficient $a_p$ of $q^p$ in the modular form

$$A(q) = q \prod_{n=1}^{\infty} (1 - q^n)(1 - q^{23n})$$

is equal to either $2$ or $0$.

The question of what happens for more general polynomials is related to the Langlands program / nonabelian class field theory. It's known that by the Chebotarev density theorem the density of primes such that $f(x) \bmod p$ has a root ("solvable" is bad terminology here, since it conflicts with "solvable" meaning that the splitting field of $f(x)$, or equivalently its Galois group, is solvable) is the density of elements of the Galois group $G$ of $f$ fixing at least one root, which in particular is at least $\frac{1}{|G|}$. (For example, if $\deg f = n$ and $G = S_n$, which is the generic case, then the density is the density of permutations with at least one fixed point which is asymptotically $1 - e^{-1}$.) But as far as I know (which is not that far, I am not a number theorist) a description of exactly which primes these are for a Galois group $G$ that doesn't embed into $GL_2(\mathbb{Z}/\ell)$ for some prime $\ell$ is out of reach. I have no idea what is even conjectured here.

For polynomials in more than one variable things are actually easier: generically we expect a polynomial $f(x_1, \dots x_n) \in \mathbb{Z}[x_1, \dots x_n]$ to have a zero $\bmod p$ for sufficiently large $p$. Heuristically the idea is that the variety $\{ f = 0 \}$ has dimension $n-1$ so we expect there to be approximately $p^{n-1}$ points on it $\bmod p$, and this intuition can be made precise using the Lang-Weil bound or (at least in the smooth irreducible case) the Weil conjectures. In the case $n = 2$, if the homogenization of $f$ defines a smooth projective curve of genus $g$ then we have the Hasse-Weil bound (implied by the Weil conjectures for curves) which implies that there are at least

$$p - 2g \sqrt{p} + 1 - \deg f.$$

zeroes of $f(x) \bmod p$ (the final term comes from the need to subtract points at infinity), which is positive for $p$ sufficiently large.

Your use of Hensel's lemma in the third paragraph is incorrect.

In general this is a fascinating topic but it doesn't have all that much relevance to the question of finding a prime $p$ such that you can show a Diophantine equation has no solution $\bmod p$, since you only need to find one such prime, not all of them. Mostly you're just going to use the fact that $x^k$ takes on $\frac{p-1}{\gcd(k, p-1)} + 1$ values $\bmod p$.


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