# Legendre symbol question

Given an integer $N \geq 2$, are there infinitely many integers $d$ such that the Legendre symbol $(\frac{d}{p}) = 1$ for all prime $p \leq N$?

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Let the primes $\le N$ be $p_0$ to $p_k$. Consider the system of congruences $d\equiv 1^2\pmod{p_i}$, $i=0$ to $k$. By the Chinese Remainder Theorem, this system has infinitely many solutions. Each is a quadratic residue modulo all the $p_i$.
Even stronger, there are infinitely many such primes $d$.... –  N. S. Jan 19 '13 at 2:13
Why restrict to $p\le N$? Just take $d=k^2$ and you have $(\frac{d}{p}) = 1$ for all primes $p$ (except for those dividing $k$).