Mystxgon
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#1
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Prove that the sum of three consecutive square numbers is one less than a multiple of 3

If u could use n² + (n+1)² + (n+2)² please
Last edited by Mystxgon; 1 month ago
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macy_m
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Okay so I’ve checked it and I believe you can, first thing expand the brackets and sum everything up then tell me what you got and I’ll check
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Mystxgon
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(Original post by macy_m)
Okay so I’ve checked it and I believe you can, first thing expand the brackets and sum everything up then tell me what you got and I’ll check
I got 3n²+6n+5 after expanding it all and adding them
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macy_m
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(Original post by Mystxgon)
I got 3n² 6n 5 after expanding it all and adding them
So then you’d factorise the 3 our and be left with a 5 outside but that wouldn’t work so you’d have to have a -1 outside, do you know how you could word it then?
Last edited by macy_m; 1 month ago
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DFranklin
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So, what do you get if you add 1? Can you see why it's a multiple of 3?

Incidentally, you could also consider your 3 consecutive squares to be (n-1)^2, n^2, (n+1)^2, in which case symmetry is going to make your calculations slightly easier. No big deal here, but something to be aware of for harder problems.
Last edited by DFranklin; 1 month ago
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Mystxgon
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(Original post by macy_m)
So then you’d factorise the 3 our and be left with a +5 outside, do you know how you could word it then?
Thank u for the help i understand it now 3(n²+2n+2) -1 i believe?
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macy_m
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(Original post by Mystxgon)
Thank u for the help i understand it now 3(n²+2n+2) -1 i believe?
Yes !
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