# Let x be an integer. Prove that if x^2 − 6x + 5 is even then x must be odd.

## Solution

### Proof.

Assume for the sake of contradiction that x^{2} − 6x + 5 is even and x is even.

By definition of even, x = 2c where c is an integer.

By substitution,

x^{2} − 6x + 5 = (2c)^{2} − 6(2c) + 5

= 4c^{2} − 12c + 5

= 4c^{2} − 12c + 4 + 1

= 2(2c^{2} − 6c + 2) + 1.

Since 2c^{2} − 6c + 2 is an integer, x^{2} − 6x + 5 must be odd, which is a contradiction.