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2 0 $begingroup$ Is there a proof that for any number $a$ , there must be at least one prime $p$ such that $(a/p)=-1$ , where $(a/p)$ is the Legendre symbol? In other words, for all $a$ , is there at least one prime $p$ such that $a$ is a quadratic nonresidue modulo $p$ ? EDIT: Due to the comments pointing out that there is no such $p$ for $a=x^2$ , my question remains the same, except for only all $aneq x^2$ . number-theory elementary-number-theory legendre-symbol share | cite | improve this question edited Dec 30 '18 at 0:45 Tejas Rao