I want to know whether there exist a nonsquare 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.



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 $(n/p) = 1$. 


This follows from the Chebotarev density theorem, or from the earlier and easier Frobenius density theorem. The polynomial $f(x):=x^2n$ is irreducible in $\mathbb{Q}[x]$, so these density theorems imply that the mod $p$ reduction of $f(x)$ is irreducible for infinitely many primes $p$ (in fact: for half of all primes $p$). It would be interesting to know a proof that didn't rely on these density theorems. 


A Chebotarevfree argument is given by our own @Pete L Clark here: http://math.stackexchange.com/questions/6976/provingthatanintegeristhenthpower 


Let $a$ be any nonsquare. Then write $a=p^nm$ for some odd $n$ and prime $p$ which does not divide $m$. By Dirichlet's Theorem on primes in arithmetic progressions and the Chinese Remainder Theorem we can find a prime $q$ which is $1\mod 4$, $(qp)=1$, and $1\mod l$ for each prime $l$ dividing $m$. Then by quadratic reciprocity, $(aq)=(pq)^n(mq)=(qp)^n=1$ (where $(\cdotl)$ is the Legendre symbol modulo $l$). 

