elementary number theory - induction proof of the fact that $n^2-9n+10$ is even

I have following problem: I need to prove that $n^2-9n+10$ is even, by induction.
I started with $n^2-9n+10 = 2k $ for some integer $k$.
$n=1$: $1-9+10 = 2$ which is even
for $n=k$ : $k^2-9k+10 = 2k $
for $n=k+1$: $(k+1)^2 - 9(k+1) +10 = 2(k+1) $
I do not know how to continue and would have liked to ask you if you could help me with this matter?
Thanks in advance!!!

Answer

When $n=1$, $1^2-9 \cdot 1 +10 = 2$ is even. Now, suppose there is some $n$ for which $n^2-9 \cdot n + 10$ is even. That is, there is a $l \in \mathbb{Z}$ with $n^2-9 \cdot n + 10=2l$. We need to show that $(n+1)^2-9 \cdot (n+1) +10$ is again even. Note:

\begin{equation}
(n+1)^2-9 \cdot (n+1) +10=n^2+2n+1-9n-9+10\\
=(n^2-9n+10)+2n-8 \\
=2l+2(n-4) \\
=2(l+n-4).
\end{equation}