I was working on my homework for Discrete Math, and we were asked to "Prove: $6 \mid n^{3}+5n$,where $n\in \mathbb{N}$" my solution varied significantly from how I have seen it done by others. I noticed a pattern and used it to say "$6\mid Q(n)$ iff $Q(k+1)-Q(k)$,where $k in \mathbb{N}$, is also divisible by $6$ ∵ $Q(k+1)-Q(k)$ represents the recursive process of mathematical induction for testing divisibility." I was not satisfied with it, even though it appears true. Attached is a picture for better context. But my overall question is has anyone heard of a proof that verifies the pattern I was describing, or could they come up with one?
It appears that when one subtracts $P(k)$ via the induction hypothesis from $P(k+1)$, then one is left with a new function $Q(n)$, which we know logically needs to be divisible by the target number for our proof to work. Therefore, $Q(k+1)-Q(k)$ will either yield a new function that is divisible by the term one is looking for, or will lead to a new function that logically needs to be divisible by the target number for our proof to work, and so on and so forth.
Nice!
Yes that's correct, if you have a sequence $a_1,a_2,a_3,\ldots$ of terms which are all divisible by a fixed integer $m$, then their differences must be also. If you are trying to prove that each of the $a_i$ are divisible by $m$, then you could construct a new sequence: \begin{align*} b_1&=a_2-a_1 \\ b_2&=a_3-a_2 \\ b_3&=a_4-a_3 \\ \vdots \end{align*}
and try to prove that the $b_i$ are divisible by $m$. Provided you prove that $m\mid a_1$ as well, this would mean that $a_n=a_1+b_1+b_2+\cdots+b_n$ is also divisible by $m$.
In your example, this works particularly nicely, because $a_n$ is a cubic polynomial in $n$. When you calculate the the difference between $a_{n+1}$ and $a_n$, you get a quadratic polynomial in $n$, from which is it simpler to prove the divisibility.
In general in fact, if you have a $k$th degree polynomial $p(x)$, and you calculate the difference
$$p(x+1)-p(x)$$
the resulting polynomial will have degree $(k-1)$ (because the $x^k$ terms cancel). Using your idea you can repeatedly calculate differences, decreasing the degree of the polynomial, until you reach a constant polynomial, from which it will be trivial to calculate any divisibility.
I don't believe this has a name, but it is a relatively standard technique when trying to inductively prove divisibility of polynomials such as this one, along with factoring.