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I'm relatively new with proofs and am trying to self-teach. I'm currently going through questions that unfortunately have no solutions... I've been doing well until I struck this one:

If l, m, and n are consecutive integers, then 12 does not divide $l^2 + m^2 + n ^2 +1$.

I know that proof by contradiction is p and (not q) => C. So to start off, I assume that it IS divisible by 12 and I have $l$, $m=l+1$, n=$l+2$. Therefore $l^2 + m^2 + n ^2 +1 = 3l^2+6l+6 = 3(l^2+2l+2)$. This is as far as a got. Any help or hints would be appreciated.

4yourneed
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    Now you have to show that 4 does not divide the expression in paranthesis – Atul Mishra Jun 25 '17 at 01:19
  • You are very close. Since $12 \mid l^2+m^2+n^2 +1$ it must be that $12 \mid 3(l^2+2l+2)$. So, $4$ must divide $l^2+2l+2$. Is there any value of $l$ to allow that? – Harry Jun 25 '17 at 01:20

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hint

Let us continue your work.

$$12|3 (l^2+2l+2)\implies $$ $$4|l^2+2l+2 \implies $$

$$2|l^2+2l+2 \implies $$ $$2|l \implies l=2p \implies $$ $$4|4p^2+4p+2 \implies 4|2$$

You can finish.

hamam_Abdallah
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  • I'm not too familiar with "|" but I assume it means X is a factor of Y? How did you get from the second to third line? Thanks! – 4yourneed Jun 25 '17 at 04:28
  • @4yourneed If $l^2+2l+2$ is divisible by 4 then surely it must also be divisible by 2, as it is even. In other words, if $4|l^2+2l+2$, then $2|l^2+2l+2$ as well. – Vedvart1 Jun 25 '17 at 09:21
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Hint:

$$ l^2+2l+2=(l+1)^2+1 $$

Tucker
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Lets try it in this way

Take $l=m-1,n=m+1$

Then the expression becomes $$(m-1)^2+m^2+(m+1)^2+1$$

Which reduces to $$3(m^2+1)$$

Now if $m$ is even then the expression in paranthesis is odd and not divisible by $4$

If $m$ is odd , then $m^2=8k+1$ and $m^2+1=8k+2$ which is not divisible by 4

So the work is done

Similar argument you can use in your expression

Hope it helps.....!!!!

Atul Mishra
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