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Question: Show that for all natural numbers $n$ which greater than or equal to 1, then 9 divides $(n-1)^3+n^3+(n+1)^3$.

Hence, $(n-1)^3+n^3+(n+1)^3 = 3n^3+6n$, then $9c = 3n^3+6n$, then $c=(n^3+2n)/3$. Therefore $c$ should be integers, but I don't know how to do it at next step ?

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$3n^3 + 6n = 3n(n^2 + 2)$ but $n^2$ is always 1mod3 –  Dinesh May 8 '11 at 9:44
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Please do not use (homework) as the sole tag for your questions; use it along with a tag pertaining to the appropriate mathematical subject. –  J. M. May 8 '11 at 10:17
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8 Answers

up vote 15 down vote accepted

To say that $3n^3+6n$ is divisible by $9$ is equivalent to saying that $n^3+2n=n(n^2+2)$ is divisible by $3$. Now split to three cases: $n=3k$ or $n=3k+1$ or $n=3k+2$ (explain why this is true.) Now you have to check just 3 cases.

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Let $\wp(n) = (n-1)^3 + n^3 + (n+1)^3$. Observe that $\wp(1) = 9$, and so is divisible by $9$. To complete this inductive proof, we want to prove that $\wp(n+1) - \wp(n)$ is divisible by $9$.

$\wp(n+1) - \wp(n) = n^3 + (n+1)^3 + (n+2)^3 - (n-1)^3 - n^3 - (n+1)^3 $

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

and so is also divisible by $9$. We have thus shown that $\wp(1) = 0 \mod 9$, and that $\wp(n+1) - \wp(n) = 0 \mod 9$, and so $\wp(2) = \wp(1) + \wp(2) - \wp(1) = 0 \mod 9$, and inductively proceed from there...

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Never seen a script p like that used as a symbol. Where does it come from? –  crasic May 8 '11 at 19:55
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@crasic: It's called Weierstrass p (LaTeX: \wp). It's almost exclusively used for W.'s $\wp$-function(s). W. introduced this symbol specifically for these doubly periodic functions. –  t.b. May 8 '11 at 20:18
    
@Theo funny that I had seen weierstrass p-functions without the script p. Is the W-p relevant in this situation (with the polynomial presented by Isaac) or is it just a stylistic variation on his part? –  crasic May 9 '11 at 0:36
    
@crasic: I would say the elliptic functions don't really come into play here and Isaac just wanted a fancy-looking "p"; besides, Weierstrass $\wp$ requires two more arguments (for a total of three). –  J. M. May 9 '11 at 2:12
    
You are correct, J.M. I recently stumbled upon it in a paper and thought that it looked exceedingly fancy. –  Isaac Solomon May 9 '11 at 2:55
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HINT $\rm\quad\displaystyle (n+1)^3 + n^3 + (n-1)^3\ =\ 3\:(n+1)\:n\:(n-1) + 9\:n\ =\ 18\:{n+1 \choose 3} - 9\:n$

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From your computations, it's enough to show that $n^3+2n=n(n^2+2)\equiv 0\pmod{3}$. If $n\equiv 0\pmod{3}$, you are done, else $n\equiv 1,2\pmod{3}$, in which case $n^2+2\equiv 0\pmod{3}$.

Reducing it to modular arithmetic may save you a lot of messy multiplication.

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Since $(n-1)^3 + n^3 + (n+1)^3 = 3n(n^2+2)$ we know that it is divisible by $3$. If we show that $n(n^2+2)$ is also divisible by $3$ then $(n-1)^3 + n^3 + (n+1)^3$ is divisible by $3^2 = 9$. There are three cases to check:

  • $0(0^2+2) \equiv 0 \cdot 2 \equiv 0 \pmod 3$
  • $1(1^2+2) \equiv 1 \cdot 0 \equiv 0 \pmod 3$
  • $2(2^2+2) \equiv 2 \cdot 0 \equiv 0 \pmod 3$
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You want to show that $$(n-1)^3+n^3+(n+1)^3 \equiv 0 \pmod 9.$$ There are $9$ cases to check,

  • $(0-1)^3+0^3+(0+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
  • $(1-1)^3+1^3+(1+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
  • $(2-1)^3+2^3+(2+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
  • $(3-1)^3+3^3+(3+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
  • $(4-1)^3+4^3+(4+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
  • $(5-1)^3+5^3+(5+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
  • $(6-1)^3+6^3+(6+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
  • $(7-1)^3+7^3+(7+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
  • $(8-1)^3+8^3+(8+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
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Strictly speaking you have 10 cases there, but $0 \equiv 9 \mod 9$ –  Henry May 8 '11 at 12:35
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Simpler, you need only check it for 3 consecutive integers since $\rm\:(x+3)^3\equiv x^3\ (mod\ 9)\:$ Indeed, looking at your calculation shows the 3-cycling, viz. $-1+0+1,\ 0+1-1,\ 1-1+0,\ \cdots\ \ $ The calculations are useful for vividly showing the cyclicity. –  Bill Dubuque May 8 '11 at 17:35
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HINT $\rm\:\ \ mod\ 9:\ (x+3)^3\equiv\:x^3\ \Rightarrow\ f(n+1)\: \equiv\ f(n)\ $ so, by induction, $\rm\ f(n)\equiv\: f(1)\equiv\: 0 $

E.g. a specific inductive step: $\rm\: mod\ 9\::\ \ 3^3\:\equiv\: 0^3\ \Rightarrow\ 1^3+ 2^3+3^3\ \equiv\ 0^3+1^3+2^3\ \equiv\ 0\:.$

NOTE $\ $ The period $3$ repeating behavior is illustrated vividly in quanta's calculations, where one can explicitly see the repeating $3$-cycle $\ -1+0+1,\ \ 0+1-1,\ \ 1-1+0,\ \cdots$

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$9|3n^3 + 6n$ , is equivalent to , $3|n^3 + 2n$ , is equivalent to $3|n^3 - n + 3n$ , is equivalent to ,

$3|n^3 - n$ ; now $n^3 - n = n(n-1)(n+1)$ is the product of three consecutive integers whenever $n$

is an integer and since consecutive multiples of $3$ occur with a difference(gap) of $3$ ,

$n(n-1)(n+1)=n^3 - n$ must be divisible by $3$ whenever $n$ is an integer.

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