# Is my proof valid for $9$ dividing sum of three consecutive cubes?

I am trying to use induction. Have I applied it correctly / rigorously enough?

Prove that the sum of three consecutive cubes are divisible by $9$.

Base case: Let $n=0$. Then $0^3 + 1^3 + 2^3 \equiv 9 \equiv 0 \bmod 9$, and so this is true.

Inductive phase: Suppose $n^3 + (n+1)^3 + (n+2)^3 \equiv 0 \bmod 9$. We want to prove that $(n+1)^3 + (n+2)^3 + (n+3)^3 \equiv 0 \bmod 9$.

Subtracting one from the other, we get $(n+3)^3 - n^3 \equiv 0 \bmod 9$.

This expands to $9 n^2+27 n+27 \equiv 0 \bmod 9$.

Divide both sides and modulus by $9$:

$n^2+3n+3 \equiv 0 \bmod 1$, which is true because $1$ divides all positive integers.

Therefore this completes the proof.

Did I miss anything? Is my proof complete? What could I do better?

• The main part of your proof is backwards. After stating the inductive hypothesis, you should start with $n^2+3n+3\equiv 0\pmod{1}$, and continue from there. You need to end with $(n+1)^3+(n+2)^3+(n+3)^3\equiv 0$, not start with it. Dec 29 '15 at 0:55
• @vadim123 Does this ordering matter? I assumed that the goal was to state some sort of link from one case to the next (since the next three consecutive cubes are created by increasing $n$ by one), and then just show that this new assumption is true, too.
– AJJ
Dec 29 '15 at 1:03
• Using the idea "we want to prove this so we manipulate it and now we want to prove this" is awkward and misleading. Simply say "we need to show ($n + 3)^3 - n^3$ is divisble by three" and leave out "= mod 9" from your statements Dec 29 '15 at 1:05
• There is another proof. For three consecutive numbers, one is $-1 \pmod 3$, another is $0 \pmod 3$, and finally one is $+1 \pmod 3$. These can come in any cyclic permutation order, but that is not important. So the sum of their cubes is congruent to $(-1)^3 + 0^3 + (+1)^3 = 0$ modulo $3$ as wanted. Dec 29 '15 at 1:39
• @JeppeStigNielsen: The OP wanted modulo 9. If the OP believes the "Freshman's Dream" (i.e., that the Frobenius map is a homomorphism), then your suggestion leads directly to a proof.
– Snow
Dec 29 '15 at 1:54

First note that $9 \equiv 0 \pmod 9$ and $27 \equiv 0 \pmod 9$, so $9n^2+27n+27 \equiv 0 \pmod 9$. Using your work, $$(n+1)^3+(n+2)^3+(n+3)^3 = n^3 + (n+1)^3+(n+2)^3 + (9n^2+27n+27).$$ But $n^3 + (n+1)^3+(n+2)^3 \equiv 0 \pmod 9$ and $9n^2+27n+27 \equiv 0 \pmod 9$, so $$(n+1)^3+(n+2)^3+(n+3)^3 \equiv 0 \pmod 9.$$