Solution modulo every $n$ may not imply solution in $\mathbb{Z}$ If we have an equation $x^2+(odd)x+(odd)=0$, then, since modulo $2$ it has no solution, thus it has no solution in $\mathbb{Z}$.
My question could be trivial; but I was unable to search if it is known.
Is there any quadratic, cubic or higher degree equation in one variable over $\mathbb{Z}$, such that it has a solution modulo every $n\ge 2$, but no solution in $\mathbb{Z}$?
 A: Let $f$ be a monic polynomial in $\Bbb Z[t]$. It follows from density theorems from algebraic number theory that if $f$ is irreducible and of degree $>1$ it cannot have a root modulo every prime, in fact the set of primes modulo which it has no root has positive density (see e.g. Lang's ANT, Corollary on p. 170).
This shows that we cannot find a monic polynomial $f$ of degree $2$ or $3$ with no roots in $\Bbb Z$ but roots mod every prime, a fortiori we cannot find one having roots modulo every $n\geq2$.
I don't know the answer for polynomials $f$ of degree $3<\deg f<8$, but here is a way to construct polynomials of degree $\ge8$ that have no root in $\Bbb Z$ but a root modulo every $n$. By the CRT it suffices to consider prime powers. Note first that the polynomial $x^2+x+2$ has a simple root mod $2$, hence a root modulo $2^m$ for all $m\geq1$ by Hensel's lemma. For all primes $p\geq3$ one of $2,7$ and $14$ will be a quadratic residue mod $p$ and thus the polynomial $(x^2-2)(x^2-7)(x^2-14)$ will have a root modulo $p^m$ for all $m\geq1$. Putting these together we see that the polynomial $$f=(x^2+x+2)(x^2-2)(x^2-7)(x^2-14)$$ has a root modulo every $n\geq2$ but no root in $\Bbb Z$.
Maybe one can find polynomials of smaller degree?
