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!!!

Expand $(k+1)^2 - 9(k+1) + 10$ to get $k^2 + 2k + 1 - 9k-9 = k^2 -9k + 10 - 18$, and use the induction hypothesis. — dannum, Oct 12 '16 at 15:40
Thank you, I will have a look at some more proofs. Thank you for your advices! — noGo, Oct 12 '16 at 15:44
This can also be proven without induction, since for odd $n$: $n^2$ is odd as is $9\cdot n$, so their difference is even, and adding 10 keeps it even.
Similarly, for even $n$, $n^2$ and $9\cdot n$ are both even, so their difference is as well. — Giuseppe, Oct 12 '16 at 15:46

score 1 · Accepted Answer · answered Oct 12 '16 at 15:41

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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}

answered Oct 12 '16 at 15:41

Sloan

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Wow, that was really fast. Thanks a lot I will do some more induction proofs to practise it more. This helps me a lot, thank you!! – noGo Oct 12 '16 at 15:43

score 1 · Answer 2 · answered Oct 12 '16 at 15:46

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Induction plus modular arithmetic:

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

answered Oct 12 '16 at 15:46

Siong Thye Goh · Answer 3 · 2016-10-12T15:52:30.543

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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}

edited Oct 12 '16 at 15:52

answered Oct 12 '16 at 15:40

Siong Thye Goh

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Thank you for your quick response Siong Thye Goh! – noGo Oct 12 '16 at 15:45

score 0 · Answer 4 · answered Oct 12 '16 at 15:43

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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. $$(n+1)^2-9(n+1)+10=(n^2-9n+10)+2n+1+10-9$$ $$(n^2-9n+10)+2(n+1)$$

answered Oct 12 '16 at 15:43

AlgorithmsX

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

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