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

Question

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 1

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}

Answer 2

Remark:

The $k$ in your current setting are used to mean two different things, it would be great to use different notations.

Suppose $k^2-9k+10=2l$ for some integer $l$.

Show that $(k+1)^2-9(k+1)+10$ is even.

\begin{align} (k+1)^2-9(k+1)+10&=k^2+2k+1-9k-9+10\\ &=(k^2-9k+10)+(2k+1-9) \end{align}

Answer 3

Just expand $(n+1)^2-9(n+1)+10$, group the terms so that you have $(n^2-9n+10)$ plus some function, and prove that that function is even. $$(n+1)^2-9(n+1)+10=(n^2-9n+10)+2n+1+10-9$$ $$(n^2-9n+10)+2(n+1)$$

Answer 4

Induction plus modular arithmetic:

If $$n^2-9n+10\equiv0\mod2,$$ then $$(n+1)^2-9(n+1)+10\equiv(n^2-9n+10)+(2n+1-9)\equiv0\mod2.$$