So im working on induction proofs but im having a bit of trouble with them. If you guys could help me that would be greatly appreciated, they're just sort of confusing to me.
Use induction to prove the following theorem:
Theorem: For each integer $n$, $n^3-n\equiv 0\pmod{6}$
$(k+1)^3-(k+1)=k^3-k+3k^2+3k=k^3-k+3k(k+1)$. You know by hypothesis that $k^3-k$ is divisible by $6$, the other term is divisible by $6$ because $k(k+1)$ is always even.
In your inductive step you write $k^3-k=(k+1)^3-(k+1)$, which I hope was meant to mean equivalence modulo $6$ rather than equality, but in any case it's important to prove, not start from, such an equivalence to establish the inductive step. The desired condition is equivalent to $6$ dividing $(k+1)^3-k^3-1=3k(k+1)$, or to $k(k+1)$ being even. This follows from the fact that $k,\,k+1$ cannot both be odd.
There is mistake in your proof. Note that $(k+1)^3-(k+1)=k^3+3k^2+3k-k=k^3-k+3k (k+1) .$ Now $k^3-k$ is a multiple of $6$ by assumption and $k (k+1) $ is a multiple of $2$(why?). Can you complete now?
Much easier to prove in steps of $6$:
Let the base case be $-2\le n\le 3$. This is proved by explicit calculation; there are only 6 cubes to compute, and only one of these gives more than a single-digit result.
For $n\ge 4$, the induction hypothesis is $(n-6)^3-(n-6)\equiv 0\pmod 6$. If we expand using the binomial theorem, this gives us $$ n^3 + (\text{terms that are multiples of $6$}) - n + (\text{yet another multiple of $6$}) \equiv 0 \pmod 6 $$ which is to say $n^2-n\equiv 0\pmod 6$ as desired.
Now if $n$ is negative, then $-n$ is certainly $\ge -2$, so we know that $(-n)^3 - (-n) \equiv 0\pmod 6$. Negating both sides of this again gives $n^2-n\equiv 0\pmod 6$.
