Step 1

a) ${a}^{2}-5a+6=(a-2)(a-3)$

Exactly one of $a-2$ or $a-3$ must be even. So the product must be even. (The product of an even and odd number is even).

- If you must, consider cases: Every integer is even or odd. If a is even, then show that so is $a-2$, and hence so is $(a-2)(a-3)$. And if a is odd, then show that $a-3$ is even, and hence so is $(a-2)(a-3)$

Step 2

For (b), consider cases: For any two integers a,b, one of the following must be true:

1. a, b both even,

2. a, b both odd,

3. a even and b odd,

4. a odd and b even.

When case (1) or case (2) holds, show that ${a}^{2}-5b$ is even. Then show that given either case (1) or case (2), ${b}^{2}-5a$ is also even.

(In case (3) and in case (4), ${a}^{2}-5b$ is odd, so those cases are irrelevant, since no claim is being made about ${b}^{2}-5b$ when ${a}^{2}-5$ is odd.)