Easy Proof

[ruud]

I'm getting confused and can't seem to wrap my head around this problem. Prove that the sum of the squares of any 3 consecutive odd numbers when divided by 12 gives a remainder of 11.

I'm not sure how to set this up or proceed I figured that

(n^2 + (n +2)^2 + (n+4)^2)/12 = x + 11


Where n is any odd integer

then I got

3n^2 +12n + 20 = 12x +123
3n(n+4) = 12x + 103
I'm not sure where to take it from here
Can someone start out on an alternate solution or set it up differently for me?

Thanks

[matt grime]

you will get a lot further if you let n=2k+1 which is always what you should do when asked to consider odd numbers


the bit where you write x+11 is wrong, as that isn't what 'has remainder 11 on dividing by 12' means.

omit the x entirely

look at the line 3n^2 +12n + 20

this has the same remainder as 3n^2 +20, now put n=2k+, and remember 12 = 0 mod 12.

[Warr]

(2n+1)^2 + (2n+3)^2 + (2n+5)^2
after expanding and simplifying
12n^2 + 36n + 35

as you see, 12 goes into 12n^2 evenly, as does 12 into 36...so far so good

and then we come to 35...35/12 = 2 + 11R....there is your 11

ie.

in general terms f[unction](n)=d[ivisor](n)q[uotient](n) + r[emainder](n)
where n is a positive integer

so the division statement would be:
12n^2 + 36n + 35 = (12)(n^2 + 3n + 2) + 11