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?
 A: $$a^2 - 5a + 6 = (a-2)(a-3) = (a-3)((a-3)+1)$$
Now can you show that one of the two factors must be even? (It would be a good exercise to establish to your satisfaction that given any two consecutive integers, exactly one of them is even.)
Sketch of proof by cases : either $a$ is odd, or $a$ is even:


*

*If $a$  is odd, then $(a-3)$ is even.

*If $a$ is even, then $(a-2)$ is even.


Since one of the factors in $(a-2)(a-3)$ is necessarily even, whatever the value of $a$, the entire product must be even. 
A: By Fermat's little theorem, $a^2\equiv a\pmod{2}$; so
$$
a^2-5a+6\equiv a-a\equiv0\pmod{2}
$$
A: $(2n)^2-5(2n)+6=4n^2-10n+6$
$(2n+1)^2-5(2n+1)+6=4n^2-6n$
Do i have to say something more?
A: This probably goes into more detail than is needed, but here it is. 
$a$ is even: If $a$ is even, then $a = 2m$, where $m\in \mathbb{Z}$. Then we have
$$
(2m)^2-5(2m)+6 = 4m^2-10m+6=2(2m^2-5m+3).
$$
Let $\eta = 2m^2-5m+3$, where $\eta \in \mathbb{Z}$. Thus, $a^2-5a+6$ is even when $a\in\mathbb{Z}$ is even.
$a$ is odd: If $a$ is odd, then $a = 2m-1$, where $m\in\mathbb{Z}$. Then we have
$$
(2m-1)^2-5(2m-1)+6=4m^2-4m+1-10m+5+6=2(2m^2-7m+6).
$$
Let $\gamma = 2m^2-7m+6$, where $\gamma\in\mathbb{Z}$. Thus, $a^2-5a+6$ is even when $a\in\mathbb{Z}$ is odd. 
Hence, $a^2-5a+6$ is even when $a\in\mathbb{Z}$ is either even or odd. 
A: I guess that by "direct proof" they mean "not by contradiction." Such a proof would start out something like "Suppose $a^2 - 5a + 6$ is odd." A direct proof, on the other hand, doesn't set up false assumptions to knock down later.
This is the way I'd go about a direct proof:
If $a$ is odd, then so is $a^2$. The difference between two odd numbers is an even number. Thus $a^2 - 5a$ is even. When you add an even number to an even number, you get another even number. So $a^2 - 5a + 6$ is even.
But if $a$ is even, then so are $a^2$ and $5a$. The difference between two even numbers is also an even number. Thus $a^2 - 5a$ is even, as well as $a^2 - 5a + 6$.
A: HINT
I want to add to amWhy's hint: 
An integer $x$ is even if there exists an integer $m$ so that $x=2m$.
An integer $x$ is odd if there exists an integer $m$ so that $x=2m+1$.
Let $a$ be be even:
$(a-2)(a-3)=(2m-2)(2m-3)=4m^2-6m-4m+6=4m^2-10m+6$
$$=2(2m^2-5m+3)$$
where we let $ \phi =(2m^2-5m+3)$, which is an integer by closure and so we have $2\phi$
This proves that when $a$ is even your entire product is even.  You can make a similar argument for odd numbers. 0 is trivial.  Hope this hint helps. 
A: 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:
$$\calc
\even{a^2 - 5 \times a + 6}
\calcop={$\ref 0$, two times}
\even{a^2} \;\equiv\; \even{5 \times a} \;\equiv\; \even 6
\calcop={$\ref 1$, two times}
\even a \lor \even a \;\equiv\; \even 5 \lor \even a \;\equiv\; \even 6
\calcop={5 is odd, 6 is even; logic: simplify}
\even a \;\equiv\; \even a
\calcop={logic: simplify}
\true
\endcalc$$
