Let's define $S=\#\lbrace n\leq x$ such that $n$ is a quadratic residue modulo $p$ for all $p\leq \sqrt {x}\rbrace$.
I have to show that $\lfloor{\sqrt{x}}\rfloor\leq S\leq C\sqrt{x}$ for a certain absolute constant $C$.
I try to solve it using Brun's method, sieving the set $A=\lbrace n\leq x\rbrace$ using $P=\displaystyle \prod_{p\leq \sqrt{x}} p$; but maybe I have to insert some more hypothesis on the prime in this product.
Can someone help me with this?