Divisibility of 9 and $(n-1)^3 + n^3 + (n+1)^3$

Question

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 ?

Answer 1

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.

Answer 2

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...

Answer 3

HINT $\rm\quad\displaystyle (n+1)^3 + n^3 + (n-1)^3\ =\ 3\:(n+1)\:n\:(n-1) + 9\:n\ =\ 18\:{n+1 \choose 3} - 9\:n$

Answer 4

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.

Answer 5

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:

• $0(0^2+2) \equiv 0 \cdot 2 \equiv 0 \pmod 3$
• $1(1^2+2) \equiv 1 \cdot 0 \equiv 0 \pmod 3$
• $2(2^2+2) \equiv 2 \cdot 0 \equiv 0 \pmod 3$

Answer 6

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

• $(0-1)^3+0^3+(0+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
• $(1-1)^3+1^3+(1+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
• $(2-1)^3+2^3+(2+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
• $(3-1)^3+3^3+(3+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
• $(4-1)^3+4^3+(4+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
• $(5-1)^3+5^3+(5+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$
• $(6-1)^3+6^3+(6+1)^3 \equiv -1+0+1 \equiv 0 \pmod 9$
• $(7-1)^3+7^3+(7+1)^3 \equiv +0+1-1 \equiv 0 \pmod 9$
• $(8-1)^3+8^3+(8+1)^3 \equiv +1-1+0 \equiv 0 \pmod 9$

Answer 7

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 $\ -1+0+1,\ \ 0+1-1,\ \ 1-1+0,\ \cdots$

Answer 8

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.