Proof By induction of $n^3+2n$, is that circular reasoning?

Question

I will skip the base step in this induction proof. We know the formula $n^3+2n$ and we assume that is divisible by $3$ , so if $(n+1)^3+2(n+1)$ is still divisible by $3$ then the value $n^3+2n$ is divisible by $3$. I was wondering if we could apply the same reasoning by using the value $n-1 $. We assume that $(n-1)^3+2(n-1)=n^3-1+3n-3n^3+2n-2$ is divisible by $3$ we can split the latter equation in $2$ parts: (part1) $n^3+2n$ and (part2) $-3n^3-3+3n$ . As we can see part2 is divisible by $3$ (everything is multiple of $3$) and part1 must be divisible by $3$ because our hypothesis says so. I was wondering if my reasoning is circular, because I assumed that $n^3+2n$ is true to derive $(n-1)^3+2(n-1).$

Answer 1

Your reasoning isn't circular, but it also doesn't prove that $n^3+2n$ is divisible for all $n$.

When proving something by induction, you prove that if it holds for $n$, then it also holds for $n+1$, and you prove the base case $n_0$. Then by the structure of the natural numbers, you have proved it for all $n\geq n_0$.

However, if you prove that if it holds for $n$, then it also holds for $n-1$, and you prove a base case $n_0$, then you have proved it for all $n\leq n_0$. So you haven't proved it for all natural numbers.

Answer 2

In this case induction is maybe overkill:

$n^3+2n=(n^3-n)+3n=n(n^2-1)+3n=\underbrace{(n-1)(n+0)(n+1)}_A-3n$

Notice that $A$ is a product of $3$ consecutive numbers so one of the factor is divisible by $3$ and so is $A$. Since $3n$ is also divisible by $3$ so is their sum.