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Prove that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ where $n$ is a nonnegative integer.

I have seen many questions on this site that contain the answer to this problem and I already know the solution, but I have yet to find offer a clear explanation that I am able to understand. Can somebody please go through this problem and explain step by step as if talking to an elementary school student how it is solved?

I can get this far:

First, show that this is true for n=0: $0^3+(0+1)^3+(0+2)^3=9$

Second, assume that this is true for n: $n^3+(n+1)^3+(n+2)^3=9k$

Third, prove that this is true for n+1: $(n+1)^3+(n+2)^3+(n+3)^3= 9k−n^3+(n+3)^3=$

This is the part that I get lost. Where do we get 9k-n3+(n+3)3? Why wouldn't it just be 9k?

Many thanks in advance for your generous help!

Martin Sleziak
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User
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    This has already been asked a few times on this site: http://math.stackexchange.com/questions/953974/proving-9-divides-n3-n13-n23, http://math.stackexchange.com/questions/510378/mathematical-induction-prove-that-9-divides-n3-n13-n23, http://math.stackexchange.com/questions/6805/simple-proof-by-induction-9-divides-n3-n13-n23, http://math.stackexchange.com/questions/512072/use-mathematical-induction-to-prove-that-9-divides-n3-n-13-n-23 – Martin Sleziak Apr 18 '15 at 09:25
  • Instead of asking again, you should comment on the answers that you do not understand to ask specific questions about particular statements that you do not grasp. Otherwise the problem will just repeat itself here. – user21820 Apr 18 '15 at 09:57

5 Answers5

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No induction required to prove that. Write the expression as: $$(n-1)^3+n^ 3+(n+1)^3, \quad n>0$$ instead, i. e. as: $$3n(n^2+2).$$ Now,

  • either $n\equiv 0\mod 3$, and $3n\equiv 0\mod 9$;
  • or $n\equiv \pm 1\mod 3$, and $n^2+2\equiv 0\mod 3$, so that $3(n^2+1)\equiv 0\mod 9$.
Bernard
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For base case $(n=0)$ : $9$ divides $0^3+1^3+2^3=9$.

For inductive step : Supposing that $\color{red}{n^3+(n+1)^3+(n+2)^3=9k}$ where $k\in\mathbb Z$, you need to prove that $(n+1)^3+(n+2)^3+(n+3)^3=9m$ where $m\in\mathbb Z$.

$$\begin{align}\color{red}{(n+1)^3+(n+2)^3}+(n+3)^3&=\color{red}{(9k-n^3)}+(n+3)^3\\&=9k-n^3+n^3+9n^2+27n+27\\&=9(k+n^2+3n+3).\end{align}$$

mathlove
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$\underline{\text{Proof by induction:}}$

First, show that this is true for $n=0$:

$0^3+(0+1)^3+(0+2)^3=9$

Second, assume that this is true for $n$:

$n^3+(n+1)^3+(n+2)^3=9k$

Third, prove that this is true for $n+1$:

$\color{red}{(n+1)^3+(n+2)^3}+(n+3)^3=$

$\color{red}{9k-n^3}+(n+3)^3=$

$9k-n^3+n^3+9n^2+27n+27=$

$9k+9n^2+27n+27=$

$9(k+n^2+3n+3)$

Please note that the assumption is used only in the part marked red.

$\underline{\text{Proof by modular-arithmetic:}}$

Consider the following cases:

  • $n\equiv0\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv 0+ 1+ 8\equiv0\pmod9$
  • $n\equiv1\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv 1+ 8+ 27\equiv0\pmod9$
  • $n\equiv2\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv 8+ 27+ 64\equiv0\pmod9$
  • $n\equiv3\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv 27+ 64+ 125\equiv0\pmod9$
  • $n\equiv4\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv 64+125+ 216\equiv0\pmod9$
  • $n\equiv5\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv125+216+ 343\equiv0\pmod9$
  • $n\equiv6\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv216+343+ 512\equiv0\pmod9$
  • $n\equiv7\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv343+512+ 729\equiv0\pmod9$
  • $n\equiv8\pmod9\implies n^3+(n+1)^3+(n+2)^3\equiv512+729+1000\equiv0\pmod9$

Please note that this method is handy only when dealing with a relatively small divisor.

barak manos
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  • First, show that this is true for n=0:

    03+(0+1)3+(0+2)3=9 Second, assume that this is true for n:

    n3+(n+1)3+(n+2)3=9k Third, prove that this is true for n+1:

    (n+1)3+(n+2)3+(n+3)3= 9k−n3+(n+3)3= This is the part that I get lost. Where are you getting 9k-n3+(n+3)3? Why wouldn't it just be 9k?

     – User Apr 18 '15 at 13:31
  • @OmarN: Look at the assumption (below the line that starts with the word "second"). If you subtract $n^3$ from each side of the equation, then you get $(n+1)^3+(n+2)^3=9k-n^3$. – barak manos Apr 18 '15 at 14:36
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We can always write $n = 3q+r$ where $q$ is a non-negative integer, and $r$ is one of $0$, $1$ and $2$ (this is just division by $3$ with remainder).

Marking all terms which are obviously divisible by $9$ in red and collecting them in a "summary term" $\color{red}{9(\ldots)}$ in the following step, we have $$\begin{align} n^3 + (n+1)^3 + (n+2)^3 &= (3q+r)^3 + (3q+r+1)^3 + (3q+r+2)^2\\ &= \left(\color{red}{(3q)^3} + \color{red}{3(3q)^2r} + \color{red}{3(3q)r^2} + r^3\right) + \\ &\phantom{=} \left(\color{red}{(3q)^3} + \color{red}{3(3q^2)(r+1)} + \color{red}{3(3q)(r+1)^2} + (r+1)^3\right) +\\ &\phantom{=} \left(\color{red}{(3q)^3} + \color{red}{3(3q^2)(r+2)} + \color{red}{3(3q)(r+2)^2} + (r+2)^3\right)\\ &= \color{red}{9(\ldots)} + r^3 + (r+1)^3 + (r+2)^3 \end{align}$$

So we see that we only have to check that $r^2+(r+1)^2+(r+2)^2$ is divisible by $9$. Now on first view this doesn't seem like a win, since this has the exact same form as the expression we started with. But remember that there are only three possible values for $r$, so we can just check directly:

  • $r=0$: $r^3 + (r+1)^3 + (r+2)^3 = 0 + 1 + 8 = 9 = 9\cdot 1\quad\checkmark$
  • $r=1$: $r^3 + (r+1)^3 + (r+2)^3 = 1 + 8 + 27 = 36 = 9\cdot 4\quad\checkmark$
  • $r=2$: $r^3 + (r+1)^3 + (r+2)^3 = 8 + 27 + 64 = 99 = 9\cdot 11\quad\checkmark$
celtschk
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W/O using induction, $$n^3+(n+1)^3+(n+2)^3=3n^3+9n^2+15n+9\equiv3(n^3-n)\pmod9$$

Now $n^3-n\equiv (n-1)n(n+1)$ is a product of three consecutive integers. Exactly one of $n-1,n,n+1$ must be divisible by $3$

Hence, $3|(n^3-n)$

lab bhattacharjee
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