text: Q:
  
  is there a negative integer which is a quadratic residue mod every prime $p\equiv 7\mod 8$
  
  Is there a negative integer $n < 0$ such that the congruence $x^2 = n\mod p$ is solvable for every prime $p\equiv 7\mod 8$?
  If we remove the negativity condition it's well known that $n = 2$ works.
  
  A:
  
  The answer is no.
  Without loss of generality, we can assume that $n$ is square-free.  Then we can rewrite $n = -q_1q_2 \ldots q_k$ or $n = -2q_1q_2\ldots q_k$, where the $q_i$ are odd primes.  Here $k \ge 1$ because we can rule out $n= -1, -2$ by hand.  Then we have
  $$
  \binom{- q_1q_2 \ldots q_n}{p}
  = \binom{-1}{p}\binom{q_1}{p}\binom{q_2}{p}\ldots\binom{q_k}{p}
  = \pm \binom{p}{q_1}\binom{p}{q_2}\ldots\binom{p}{q_k}\\
  \binom{- 2q_1q_2 \ldots q_n}{p}
  = \binom{-1}{p}\binom{2}{p}\binom{q_1}{p}\binom{q_2}{p}\ldots\binom{q_k}{p}
  = \pm \binom{p}{q_1}\binom{p}{q_2}\ldots\binom{p}{q_k}
  $$
  The last equality for both lines uses quadratic reciprocity.  In both cases the $\pm$ sign is independent of $p$, and will be positive if and only if there are an odd number of $q$ of the form $4m+3$.
  Note that by Dirichlet's theorem we can choose infinitely many $p$ such that 
  $$p\equiv 7 \pmod{8}\\ p \equiv - 1 \pmod{q_i} \text{ for } i = 1, \ldots, k$$
  The computation of $\binom{n}{p}$ will then have a term of $-1$ for each $q_i$ of the form $4k+3$, along with a $-1$ in front in the case that there are an even number of such $q_i$.  Thus $\binom{n}{p} = -1$ as desired.
  
  