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I was doing a maths mock exam just today, and I found one question of a type I would normally not find difficult - well difficult. I came here due to it being a GCSE level question that I assume most would find ridiculously easy....

The question being: 'Prove that the sum of the squares of any three consecutive odd numbers is always 11 more than a multiple of 12'

So, I write out - being a non-calculator test:

$$ (2n+1)^2+(2n+3)^2+(2n+5)^2 $$ I expand the brackets and expect something nice... I get $$ (2n+1)^2+(2n+3)^2+(2n+5)^2 = 12n^2+36n+35 $$ Now, I think I know that

11 more than a multiple of 12

Means: $$ 12n+11 $$ But $$ 12n^2+36n+35 \neq 12n+11 $$

I don't know if I'm missing something fundemental here, whether I'm not seeing it, I mean the output does look similar in that 35 is 1 less than 36 in '36n' and there's a 12 there, but they're just not equal? How do? Thanks.

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    $\begingroup$ Note: $12n^2+36n+35= 12\times (n^2+3n+2)+11$. $\endgroup$ – lulu Apr 24 '17 at 16:16
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$$ (2k+1)^2+(2k+3)^2+(2k+5)^2 = 12k^2+36k+24+11 \\=12(k^2+3k+2)+11$$ $$=12n+11\tag{$n=k^2+3k+2$}$$

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note that $$12n^2+36n+35=12n^2+36n+36-1$$

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$$12n^2+36n+35$$ if you do $35-11$, you get $24$ which is a factor of $12$, allowing everything to be factorised. so $$12n^2+36n+35= 12n^2+36n+24+11 = 12(n^2+36n+2)+11$$ So the number would always have a remainder of $11$ afterwards, because we have already multiplied it by $12$ to make it a multiple of $12$.

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