How to solve this if ^denotes power ?
Hello,
I haven't found another method yet.
Do it by using the possible congruences of x to 13.
$\displaystyle x \equiv 0 [13] \Longrightarrow x^2+1 \equiv 1 [13] \not \equiv 0 [13]$
$\displaystyle x \equiv 1 [13] \Longrightarrow x^2+1 \equiv 2 [13] \not \equiv 0 [13]$
$\displaystyle x \equiv 2 [13] \Longrightarrow x^2+1 \equiv 5 [13] \not \equiv 0 [13]$
$\displaystyle x \equiv 3 [13] \Longrightarrow x^2+1 \equiv 10 [13] \not \equiv 0 [13]$
$\displaystyle x \equiv 4 [13] \Longrightarrow x^2+1 \equiv 4 [13] \not \equiv 0 [13]$
$\displaystyle x \equiv 5 [13] \Longrightarrow x^2+1 \equiv 0 [13]$
So it's 5
That is, assuming that you meant 13 divides x²+1... Your writing is not clear ~