prove that exist $a$ s.t. $(\frac{a}{m}) = -1$

Question

Let $m$ be an odd number s.t. $m = p_1\cdots p_l$ all different primes. Prove that exists a number $a$ s.t. $(\frac{a}{m}) = -1$, $(\frac{a}{m})$ being Jacobi's symbol.

I tried going over all cases of $m \pmod 8$, but I got stuck on $m\equiv 1 \pmod 8.$

Answer 1

Use the Chinese Remainder theorem to find $a\equiv b\pmod{p_1}$ and $a\equiv1\pmod{p_k}$ for $k\ge2$ where $b$ is your favourite quadratic non-residue modulo $p_1$.