Why is $x^2 + 1$ not divisible by $11$ where $x$ is an integer.
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https://math.stackexchange.com/questions/122048/1-is-a-quadratic-residue-modulo-p-if-and-only-if-p-equiv-1-pmod4 – lab bhattacharjee Mar 26 '18 at 11:57
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You can verify by hand that -1=10 is not a square modulo 11: $$ 1^2=1 \\ 2^2=4 \\ 3^2=9 =-2 \\ 4^2=16=5 \\ 5^2=25=3 \\ 6^2=36=3 \\ 7^2=49=5 \\ 8^2=64=9 \\ 9^2=81=4 \\ 10^2=100=1 \\ 11^2=0^2=0 \\ $$ . If there was such an x as described in your question, then its square would be -1 modulo 11.
Leon Hendrian
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It is special case of this theorem:
If $p\equiv 3 \pmod 4$ and $p\mid a^2+b^2$ then $p\mid a$ and $p\mid b$.
In your case: if $11\mid x^2+1$ then $11\mid 1$, a contradiction.
nonuser
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