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I have the solution for this but I get lost at the end, here's what I have so far.

basis $n = 0$; $9 \mid 0^3 + (0 + 1)^3 + (0 + 2)^2 ?$
$9 \mid 1 + 8$ = true

Induction: Assume $n^3 + (n + 1)^3 + (n + 2)^3 = k * 9$ // Why set it equal to $k * 9$? I know it works but why not just make the assumption => $n^3 + (n + 1)^3 + (n + 2)^3$ for some $n = k \ge 0$

Then, //and here's where I get lost

$(n + 1)^3 + (n + 2) + (n + 3)^3 = k * 9 + [(n + 3)^3 - n^3] = 9 (k + n^2 + 3n + 3)$
.

I've done similar examples but none like this. What am I not seeing?

Arash
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Ceelos
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  • http://math.stackexchange.com/questions/510378/mathematical-induction-prove-that-9-divides-n3-n13-n23 – njguliyev Oct 02 '13 at 10:51
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    Why induction? $$(n-1)^3+n^3+(n+1)^3=3n^3+6n=3(n^3-n)+9n=3\underbrace{(n-1)n(n+1)}_{\text{ product of }3 \text{ consecutive integers}}+9n$$ We know exactly one of $3$ consecutive integers is divisible by $3$ – lab bhattacharjee Oct 02 '13 at 12:03
  • @labbhattacharjee the problem asks for that procedure – Ceelos Oct 02 '13 at 12:10

3 Answers3

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You assume $\;9\mid\left[n^3+(n+1)^3+(n+2)^3\right]\;$ , and now you want to prove that also

$$9\mid\left[(n+1)^3+(n+2)^3+(n+3)^3\right]\;\;,\;\;\text{but}$$

$$(n+1)^3+(n+2)^3+(n+3)^3=3n^3+18n^2+42n+36=$$

$$=\left(3n^3+9n^2+15n+9\right)+9n^2+27n+27$$

and you can see the left part between the parentheses is divisible by nine by the inductive hypothesis, whereas the second part is obviously divisible by nine, too...

DonAntonio
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Set $f(n) = n^3 + (n+1)^3 + (n+2)^3$.

Hint: $f(n+1)-f(n) = (n+3)^3 - n^3 = 3 * 3 * n^2 + 3 * 3^2 * n + 3^3$.

Marc Palm
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We will show that $(n+1)^3+(n+2)^3+(n+3)^3$ is divisible by $9$. The following compuation shows that it is indeed so: $(n+1)^3+(n+2)^3+(n+3)^3= n^3+(n+1)^3+(n+2)^3+9(n^2+3n+3)$

since the sum of first 3 terms on the RHS is divisible by $9$ by assumption.

daulomb
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