Discrete Math Proof Method 
*

*Give a direct proof of the fact that $a^2-5a+6$ is even for any integer $a$.

*Suppose $a$ and $b$ are integers and $a^2-5b$ is even. Prove that $b^2-5a$  is even.
 A: For $(a)$:
$$a^2 - 5a+6 = (a-2)(a-3)$$ 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 $(a-2)(a-3)$



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


*

*$a, b$ both even,

*$a, b$ both odd,

*$a$ even and $b$ odd,

*$a$ 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.)
A: You did not indicate what you already tried, where you got stuck, or what ideas you had about this problem.  But since there is already a full answer, let me complement that with a 'simpler' one.  With that I mean that it is not required to invent a factorization up front, or to distinguish separate cases: there are no rabbits, just straightforward calculation and simplification.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\even}[1]{#1\text{ is even}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$What is required, is knowledge of the rules of logic.  Specifically, I will use the fact that logical equivalence ($\;\equiv\;$) is associative: $\;P \equiv (Q \equiv R)\;$ is equivalent to $\;(P \equiv Q) \equiv R\;$, and therefore we can just write $\;P \equiv Q \equiv R\;$.
And what rules do we know about expressions of the form $\;\even\dots\;$?  Well, we know
\begin{align}
\tag {1a}
\even{a + b} & \;\equiv\; \even a \equiv \even b \\
\tag {1b}
\even{a - b} & \;\equiv\; \even a \equiv \even b \\
\tag 2
\even{ab} & \;\equiv\; \even a \lor \even b \\
\end{align}
Please take some time to make sure you understand these fully.  And note how $\ref{1b}$ immediately follows from $\ref{1a}$ by the additional rule $\;\even{(-a)} \;\equiv\; \even a\;$.

To prove (a) we can now simply calculate
$$\calc
\even{a^2 - 5a + 6}
\calcop={rules $\ref{1a}$ and $\ref{1b}$}
\even{a^2} \;\equiv\; \even{5a} \;\equiv\; \even 6
\calcop={rule $\ref 2$, twice}
\even a \lor \even a \;\equiv\; \even 5 \lor \even a \;\equiv\; \even 6
\calcop={logic: simplify; use facts about 5 and 6}
\even a \;\equiv\; \false \lor \even a \;\equiv\; \true
\calcop={logic: simplify a bit more, using $\;\false \lor P \equiv P\;$ and $\;P \equiv P \equiv \true\;$}
\true
\endcalc$$

For (b), what can we do with the assumption that $\;\even{a^2-5b}\;$?  Let's again calculate:
$$\calc
\tag{*} \even{a^2-5b}
\calcop={rule $\ref{1b}$; rule $\ref 2$, twice}
\even a \lor \even a \;\equiv\; \even 5 \lor \even b
\calcop={use fact about 5; logic: simplify}
\tag{**} \even a \;\equiv\; \even b
\endcalc$$
And since the result $\ref{**}$ is symmetric in $\;a\;$ and $\;b\;$ --in other words: it remains the same if $\;a\;$ and $\;b\;$ are exchanged--, this proves that the assumption $\ref *$ is as well, in other words, we've proven
$$
\even{a^2-5b} \;\equiv\; \even{b^2-5a}
$$
