Show that a number divides

Question

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?

Answer 1

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?

Answer 2

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.

Answer 3

$$n^3+(n+1)^3+(n+2)^3=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8=\\ 3n^3+9n^2+15n+9=3(n^3+3n^2+5n+3)$$ 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 $3$).

Answer 4

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 $9$.