# Algebraic Proof: Sum of Squares of 3 consecutive odd numbers = 12n+11

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.

• Note: $12n^2+36n+35= 12\times (n^2+3n+2)+11$. – lulu Apr 24 '17 at 16:16

$$(2k+1)^2+(2k+3)^2+(2k+5)^2 = 12k^2+36k+24+11 \\=12(k^2+3k+2)+11$$ $$=12n+11\tag{n=k^2+3k+2}$$
note that $$12n^2+36n+35=12n^2+36n+36-1$$
$$12n^2+36n+35$$ 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 $12$.