I want to show that if $n$ is a positive integer greater than $3$, then $n^3+(n+1)^3+(n+2)^3$ is divisible by $9$. By expanding the expression and combining the like terms, I get $n^3+(n+1)^3+(n+2)^3=3(n+1)\left[(n+1)^{2}+2\right]$. The factor $3$ is easy to see. I wonder how to get the other factor $3$. Thanks.
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1Consider three cases $n=3k, 3k+1, 3k-1$. – Zuriel Oct 12 '19 at 03:19
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If one writes $m=n+1$, then $m(m^2+2)\equiv m(m^2-1)=m(m+1)(m-1)\pmod3$. – Angina Seng Oct 12 '19 at 03:21
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Alternatively, Newton's interpolation formula gives $$ n^3+(n+1)^3+(n+2)^3 = 9 \binom{n}{0} + 27 \binom{n}{1} + 36 \binom{n}{2} + 18 \binom{n}{3} $$ Yet another way, by induction:
Let $f(n)=n^3+(n+1)^3+(n+2)^3$. Then $f(n+1)-f(n)=(n+3)^3-n^3=9 n^2 + 27 n + 27$. The result follows by induction because $f(0)=9$.
lhf
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Already remarked in this dupe nine years ago. Please search first for obvious FAQs – Bill Dubuque Oct 12 '19 at 03:34
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@BillDubuque and yet, if one knows or can formulate a result without needing to look it up, there is not much reason to search elsewhere to find a reference for it... – abiessu Oct 12 '19 at 03:49
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1Rampant duplication is very bad for the site - see meta discussions. – Bill Dubuque Oct 12 '19 at 03:52
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Integers cubed $\bmod 9$ are $0,1,-1,0,1,-1,0,1,-1$ and that pattern 'loops'. Adding any three consecutive cubes gives $0 \bmod 9$
Keith Backman
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