I am to prove this statement:
For all integers n, n^2-n+3 is odd.
This is what I did, I just want to make sure its logical.
Suppose n is an integer such that n^2-n+3 is odd. this means there exists an integer, a, such that 2a+1 = n^2-n+3. It then follows that 2a = n^2-n+2 which is, by definition even. Since n^2-n+2 and n^2-n+3 are consecutive integers then it follows that n^2-n+3 is odd.