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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Easy Proof

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