how to slove the congruence such that
$\displaystyle
x^{2}+1\equiv 0 (\bmod 5)
$
cheers!
Since you're working in mod 5 you just need to check the values 0,1,2,3,4.
For 0 we have, $\displaystyle 0^2 + 1 = 1$
For 1 we have, $\displaystyle 1^2 + 1 = 2$
For 2 we have, $\displaystyle 2^2 + 1 = 5 = 0$ mod 5
For 3 we have, $\displaystyle 3^2 + 1 = 10 = 0$ mod 5
For 4 we have, $\displaystyle 4^2 + 1 = 17 = 2$ mod 5
So x = 2 or 3.
Do you understand what mod 5 means?
I think you've mistyped there. Should be 'so x = -1,1'
But -1 = 4 mod 5 so the answer is x = 1 and 4.
Test this by putting the values into $\displaystyle x^2 - 1$.
x=0 gives 0-1 = -1 = 4 mod 5 (since -1 + 5 = 5)
x=1 gives 1-1 = 0.
x=2 gives 4-1 = 3.
x=3 gives 9-1 = 8 = 3 mod 5 (since 8 - 3 = 5)
x=4 gives 16-1 = 15 = 0 mod 5 (since 15 - 5 - 5 - 5 = 0)
This confirms that 1 and 4 are the correct answers.