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How do I show that for all integers $n$, $n^3+(n+1)^3+(n+2)^3$ is a multiple of $9$? Do I use induction for showing this? If not what do I use and how? And is this question asking me to prove it or show it? How do I show it?

Matthew Conroy
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Adam
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    In most mathematical contexts, proving and showing are synonyms. You don't state the question, so there is no way for us to know that the question is asking. Have you tried anything? Why not try using induction? – Matthew Conroy Apr 07 '14 at 19:58
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    Induction is probably the best way to proceed. When $n = 1$, the expression evaluates to $9$, so the base case is clear. Now assume it holds for some $n$ and show this implies it holds for $n+1$. – Vikram Saraph Apr 07 '14 at 20:02

4 Answers4

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Here are a some ideas to work on.

First the cubes mod $9$ turn out to be $0^3=0, 1^3=1, 2^3=-1, 3^3=0, 4^3=1, 5^3=-1 \dots$

The pattern persists because $(3n\pm1)^3=27n^3\pm 27n^2+9n\pm 1\equiv \pm 1, (3n)^3=27n^3\equiv 0$ - the sum of any consecutive three is equal to $0$ mod $9$.

Another way of doing it is to set $n=(m-1)$ and note that (using the binomial expansion) $$(m-1)^3+m^3+(m+1)^3 =3m^3+6m=3m(m^2+2)$$

Modulo $3, m(m^2+2)\equiv m(m^2+3m+2)=m(m+1)(m+2)$ the product of three consecutive numbers, which is therefore divisible by $3$.

Or induction will do it because $(n+3)^3-n^3=9n^2+27n+27$ is divisible by $9$.

Mark Bennet
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  • There are three techniques here: look for a pattern and prove it; simplify the expression algebraically and show that it works; take the difference between one expression and the next and work on the difference between them as the inductive step. – Mark Bennet Apr 07 '14 at 20:18
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Well $$n^3+(n+1)^3+(n+2)^3=n^3+(n^3+3n^2+3n+1)+(n^3+6n^2+12n+8)=3n^3+9n^2+15n+9=3(n^3+3n^2+5n+3)=3(n+1)(n^2+2n+3)$$ Suppose $n=3k$ or $n=3k+1$. Then $$n^2+2n+3=3(3k^2+2k+1)$$ or $$n^2+2n+3=9k^2+6k+1+6k+2+3=9k^2+12k+6=3(3k^2+4k+2)$$ Can you take it from here?

user140943
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Use the identity: $a^3 + b^3 + c^3 = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) + 3abc$,

and substituting $a = n$, $b = n + 1$, and $c = n + 2$ into the equation to get:

$n^3 + (n+1)^3 + (n+2)^3 = (3n + 3)(n^2 + (n+1)^2 + (n+2)^2 - n(n+1) - (n+1)(n+2) - n(n+2)) + 3n(n+1)(n+2) = 3(n+1)(n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 - n^2 - n - n^2 - 3n - 2 - n^2 - 2n) + 3n(n+1)(n+2) = 3(n+1)(3n+3) + 3n(n+1)(n+2) = 9(n+1)^2 + 3n(n+1)(n+2)$. Observe that $n(n+1)(n+2$) is divisible by $3! = 6$. So $3n(n+1)(n+2)$ is divisible by $18$ and inturn divisible by $9$. So we're done.

DeepSea
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$$n^3+(n+1)^3+(n+2)^3=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8=\\ 3n^3+9n^2+15n+9=3(n^3+3n^2+5n+3)$$ Thus, this amounts to showing that $n^3+3n^2+5n+3$ is divisible by $3$ for all $n$. Obviousl, the terms $3n^2,3$ are divisible by $3$. Thus, we need to check that $\forall n,n^3+5n$ is divisible by $3$. Let's proceed by induction, letting the statement $\forall n,n^3+5n$ is divisible by $3$ be the induction hypothesis.

It's true for $n=1$. Suppose it's true for $n\geq1$. Then, $(n+1)^3+5(n+1)=(n^3+5n)+(3n^2+3n+6)$. Use now the induction hypothesis ($\forall n,n^3+5n$ is divisible by $3$).