TITLE: nth powers modulo all primes
QUESTION [15 upvotes]: Let $a \in \mathbb{Z}$, $n \in \mathbb{N}^*$ be integers, and set $P=X^n - a$. Let us consider the three following statements :
1) $P$ has a root in $\mathbb{Z}$ (i.e. $a$ is an nth power)
2) $P$ has a root in $\mathbb{Z}/d\mathbb{Z}$ for all $d \ge 1$ (i.e. $a$ is an nth power mod $d$ for all $d$)
3) $P$ has a root in $\mathbb{Z}/p\mathbb{Z}$ for all prime $p$ (i.e. $a$ is an nth power mod $p$ for all prime $p$)
I would like to prove that they are equivalent, but I still can't prove $3 \Rightarrow 1$.
Here's what I already know :

Obviously we have $1 \Rightarrow 2 \Rightarrow 3$
Proving $2 \Rightarrow 1$ is not very difficult ($2 \Rightarrow$ all $p$-adic valuations of a are multiples of $n$)
I think I can prove $3 \Rightarrow 1$ in the case $n = 2$. But my proof seems quite involved (and frankly a bit of an overkill) for this question : Roughly, if $a$ is not a square, I use the quadratic reciprocity law and Dirichlet's theorem to prove that there exist a prime $p$ for which the Legendre symbol of $a$ mod $p$ is $-1$.

So I guess my questions are : Can we prove $3 \Rightarrow 1$ ? Is there an "elementary" proof ? Thanks in advance for any comment or answer.

REPLY [4 votes]: Also, the simplest example of a polynomial that is irreducible over $\mathbb Q$ but reducible modulo every prime is the traditional fave, $x^4+1$.
Edit: as suggested by a comment (thx! @Pierre-Yves G.), one proof of this begins by noting that $x^4+1$ is the 8th cyclotomic polynomial. Since $\mathbb F_{p^2}^\times$ is cyclic, there will be an 8th root of unity in $\mathbb F_{p^2}$ exactly when $8|p^2-1$, which would mean that $x^4+1$ factors into at worst quadratic factors over $\mathbb F_p$. Since $\mathbb Z/8^\times$ is a 2,2 group, $p^2=1$ mod $8$ for every odd prime $p$. ($x^4+1=(x+1)^2$ at $p=2$.)