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Find a natural number n such that whether or not 11 is a square modulo a prime $p$ only depends on the congruence class of $p$ modulo $n$ (apart from finitely many exceptions),

and find those congruence classes

$p\equiv a(mod n)$ for which $(\frac{11}{p})= 1$

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By quadratic reciprocity, $(\frac {11}p)=(\frac p{11})$ if $p\equiv 1\pmod 4$ and $(\frac {11}p)=-(\frac p{11})$ if $p\equiv 3\pmod 4$. Thus if you know $p\bmod 44$, you know both $p\bmod 11$ and $p\bmod 4$, which is enough to determine $(\frac {11}p)$. For the list of congruence classes, play around with the concrete values.

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