Properties of numbers which are $x^2 + 1$

Question

Why is $x^2 + 1$ not divisible by $11$ where $x$ is an integer.

Answer 1

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.

Answer 2

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.