Give a direct proof of the fact that $a^2-5a+6$ is even for any $a \in \mathbb Z$ Give a direct proof of the fact that $a^2-5a+6$ is even for any $a \in \mathbb Z.$
I know this is true because any even number that is squared will be even, is it also true than any even number multiplied by 5 will be even?? is this direct proof enough?

Here is a 'logical' way to prove this, without using modular arithmetic, and without case distinctions.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\newcommand{\even}[1]{#1\text{ is even}}
$I'm assuming that we can use the following two rules:
\begin{align}
\tag 0 \even{n + m} \;\equiv\; \even n \equiv \even m \\
\tag 1 \even{n \times m} \;\equiv\; \even n \lor \even m \\
\end{align}
Now we simply calculate:
$$\calc
\even{a^2 - 5 \times a + 6}
\calcop={$\ref 0$, two times}
\even{a^2} \;\equiv\; \even{5 \times a} \;\equiv\; \even 6
\calcop={$\ref 1$, two times}
\even a \lor \even a \;\equiv\; \even 5 \lor \even a \;\equiv\; \even 6
\calcop={5 is odd, 6 is even; logic: simplify}
\even a \;\equiv\; \even a
\calcop={logic: simplify}
\true
\endcalc$$
