Diophantine Equations – Equation with No Integer Solutions but Solutions Modulo Every Integer

diophantine equationsnt.number-theoryreference-request

It's probably common knowledge that there are Diophantine equations which do not admit any solutions in the integers, but which admit solutions modulo $n$ for every $n$. This fact is stated, for example, in Dummit and Foote (p. 246 of the 3rd edition), where it is also claimed that an example is given by the equation
$$ 3x^3 + 4y^3 + 5z^3 = 0. $$
However, D&F say that it's "extremely hard to verify" that this equation has the desired property, and no reference is given as to where one can find such a verification.

So my question is: Does anyone know of a readable reference that proves this claim (either for the above equation or for others)? I haven't had much luck finding one.

Best Answer

It is actually quite straightforward to write down examples in one variable where this occurs. For example, the Diophantine equation $(x^2 - 2)(x^2 - 3)(x^2 - 6) = 0$ has this property: for any prime $p$, at least one of $2, 3, 6$ must be a quadratic residue, so there is a solution $\bmod p$, and by Hensel's lemma (which has to be applied slightly differently when $p = 2$) there is a solution $\bmod p^n$ for any $n$. We conclude by CRT. (Edit: As Fedor says, there are problems at $2$. We can correct this by using, for example, $(x^2 - 2)(x^2 - 17)(x^2 - 34)$.)

Hilbert wrote down a family of quartics with the same property. There are no (monic) cubics or quadratics with this property: if a monic polynomial $f(x) \in \mathbb{Z}[x]$ with $\deg f \le 3$ is irreducible over $\mathbb{Z}$ (which is equivalent to not having an integer solution), then by the Frobenius density theorem there are infinitely many primes $p$ such that $f(x)$ is irreducible $\bmod p$.

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