# Discrete Math Proof Method. 1. Give a direct proof of the fact that a^2-5a+6 is even for any integer a. 2. Suppose a and b are integers and a^2-5b is even. Prove that b^2-5a is even.

Discrete Math Proof Method
1. Give a direct proof of the fact that ${a}^{2}-5a+6$ is even for any integer a.
2. Suppose a and b are integers and ${a}^{2}-5b$ is even. Prove that ${b}^{2}-5a$ is even.
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Monica Dennis
Step 1
a) ${a}^{2}-5a+6=\left(a-2\right)\left(a-3\right)$
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 $\left(a-2\right)\left(a-3\right)$. And if a is odd, then show that $a-3$ is even, and hence so is $\left(a-2\right)\left(a-3\right)$
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.)

Libby Owens
Step 1
What is required, is knowledge of the rules of logic. Specifically, I will use the fact that logical equivalence ($\phantom{\rule{thickmathspace}{0ex}}\equiv \phantom{\rule{thickmathspace}{0ex}}$) is associative: $\phantom{\rule{thickmathspace}{0ex}}P\equiv \left(Q\equiv R\right)\phantom{\rule{thickmathspace}{0ex}}$ is equivalent to $\phantom{\rule{thickmathspace}{0ex}}\left(P\equiv Q\right)\equiv R\phantom{\rule{thickmathspace}{0ex}}$, and therefore we can just write $\phantom{\rule{thickmathspace}{0ex}}P\equiv Q\equiv R\phantom{\rule{thickmathspace}{0ex}}$.
Step 2
And what rules do we know about expressions of the form … is even? Well, we know

Please take some time to make sure you understand these fully. And note how (1b) immediately follows from (1a) by the additional rule
Step 2
To prove (a) we can now simply calculate

For (b), what can we do with the assumption that ? Let's again calculate:

And since the result (**) is symmetric in a and b --in other words: it remains the same if a and b are exchanged--, this proves that the assumption (*) is as well, in other words, we've proven