I want to know whether there exist a non-square number $n$ which is the quadratic residue of every prime.
I know it is very elementary, and I think those kind of number are not exist, but I don't know  

how to prove.



Does there exist a non-square number which is the quadratic residue of every prime?
This is actually an elementary consequence of quadratic reciprocity,
generalizing the familiar proof à la Euclid that that are infinitely many
primes of the form $4k+3$ 
(i.e. primes of which $-1$ is not a quadratic residue).
We want to show that there exists $p$ such that $(n/p) = -1$.
[As stated Paul's question asks only for $(n/p) \neq +1$, but this is
trivial (if $n = -1$, take $p=3$; else let $p$ be a factor of $n$), 
so we'll exclude the finitely many prime factors $p$ of $n$.]
By QR there exists a nontrivial homomorphism 
$\chi: ({\bf Z} / 4n{\bf Z})^\* \rightarrow \lbrace 1, -1 \rbrace$
such that $(n/p) = \chi(p)$ for all primes $p \nmid 2n$.
Let $a$ be *any* positive integer coprime to $4n$ such that $\chi(a) = -1$.
Then we have a prime factorization $a = \prod\_j p\_j$,
and $\prod\_j \chi(p\_j) = \chi(a) = -1$.
Therefore $\chi(p\_j) = -1$ for some $j$, **QED**.


As in Euclid we can iterate this argument to construct
infinitely many distinct $p$ for which $(n/p) = -1$.


