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.



Diophantine equation with no integer solutions, but with solutions modulo every integer
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 $\bmod p$.


