Can a number be a quadratic residue modulo all prime that do not divide it
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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
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edited Dec 30 '18 at 0:45
Tejas Rao
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