prove that if 'n' is an integer then prove that n^2 -n+2 is even

I tried mathematical induction to prove it
I used s of k = k^2-k+2
s of k+1=(K+1)^2-(k+1)+2=k^2+2k+1-k-1+2
=k^2k+k+2
=k(k+1)+2

I don't know how to proceed after this.

If you look at k, and k+1, they are consecutive numbers, one of which has to be even. Any number multiplied by an even number is even as well.

I won't bother showing it's true for 1

If k^{2}-k+2 is divisible by 2, then

(k+1)^{2}-(k+1)+2=k^{2}+2k+1-k-1+2=(k^{2}-k+2)+2k.

Since k^{2}-k+2 is divisible by 2 and 2k is divisible by 2, then

(k+1)^{2}-(k+1)+2 is divisible by 2.

If k is odd, k^2 must be odd, so k^2-k is even.
If k is even, k^2 is even, and so is k^2-k.