Wherever used, $P_n$ will here represent a polynomial with $\deg(P_n)=n$.
While working on a related area, I stumbled across a remarkable observation that, for any quadratic $P_2$; $$P_2(x)-3P_2(x+1)+3P_2(x+2)-P_2(x+3)\equiv 0$$
This is very easily proven by expansion. (Also notice the corollary $\frac{P_2(x+3\alpha)-P_2(x)}{P_2(x+2\alpha)-P_2(x+\alpha)}=3$) I immediately noticed it's similarity to the expansion of $(x-1)^3$. Written in sum notation, I have that $$\sum_{k=0}^3\binom{3}{k}(-1)^kP_2(x+k)=0$$
As a next step, I checked whether similar holds for linear functions, that is: $$P_1(x)-2P_2(x+1)+P_2(x+2)=0$$ this is true at a glance when we rearrange to $$P(x+1)=\frac{P(x)+P(x+2)}{2}$$ and whether similar holds for cubics:
$$P_3(x)-4P_3(x+1)+6P_3(x+2)-4P_3(x+3)+P_3(x+4)=0$$ by expansion, this also holds. (It's much more efficient to evaluate this if you let $x$ be the middle term).
Can I prove that: $$\sum_{k=0}^n(-1)^k\binom{n}{k}P_{n-1}(x+k)\equiv 0$$ is generally true?
Also, I did some hunting around and couldn't find this anywhere. I doubt it's fresh, so has anyone seen this before?
I proved not only this true, but the stronger case: $$\forall m\in\Bbb N_0, n\in \Bbb N, m<n; \sum_{k=0}^n(-1)^k\binom{n}{k}P_{m}(x+k)\equiv 0 $$
We begin by evaluating $$P_m(x+k)=\sum_{j=0}^mc_j(x+k)^j$$ With this, we rearrange the double sum: $$\sum_{j=0}^{m}c_j\sum_{k=0}^n(-1)^k\binom{n}{k}(x+k)^j$$ and prove the right sum is $0$ for every $j<n$
Base Case, $j=0$, we get $$\sum_{k=0}^n(-1)^k\binom nk$$
Let's prove this is $0$. We use the binomial theorem on $((-1)+1)^n$
$$0=((-1)+1)^n=\sum_{k=0}^n\binom nk (-1)^k(1)^{n-k}=\sum_{k=0}^n\binom nk (-1)^k \text{ a.r.}$$
Assumption step. We assume that: $$S_N(n)=\sum_{k=0}^n(-1)^k\binom nk(x+k)^N=0$$ holds for any $n>N+1$. (recall that our conjecture is that it holds for any $m<n$, thusly $N+1<n$ here)
Inductive step. We multiply the summand by $(x+k)$. We can separate this operation: $$S_{N+1}(n)=\sum_{k=0}^n(-1)^k\binom nk(x+k)^{N+1}=x\sum_{k=0}^n(-1)^k\binom nk(x+k)^N+\sum_{k=0}^nk(-1)^k\binom nk(x+k)^N$$ $$=xS_N(n)+\sum_{k=0}^nk(-1)^k\binom nk(x+k)^N$$ $xS_N(n)$ is trivially zero. When evaluating the Right Sum (I'll call it $S$), we note $$\sum_{k=0}^nk\binom nk =n\sum_{k=1}^n \binom{n-1}{k-1}$$ applying this identity and shifting the base by $1$ yields: $$S=n\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}(x+k)^N=nS_N(n-1)=0 \text{ a.r.}$$ under the conditions of the assumption step. Thus the desired result is proven. However, we must show that it only holds for $m<n$, that is we can only perform our inductive step $n-1$ times from the start.
Observing that the result used, more broadly stated as $$\sum_{k=0}^nk^T\binom nk=\frac{n!}{(n-T)!}\sum_{k=0}^{n-T}\binom{n-T}{k}$$ for the $T$th use (my proof uses $T=1$), only makes sense where $T\leq n$, we see why our conjecture is true only for $m<n$. Incidentally: $$\sum_{k=0}^n(-1)^k\binom nk P_n(x+k)=(-1)^nn!$$ is seen when $T=n$.
Thusly, it has been proven that: $$\forall m\in\Bbb N_0, n\in \Bbb N, m<n; \sum_{k=0}^n(-1)^k\binom{n}{k}P_{m}(x+k)\equiv 0 $$
This result is fascinating to me. I would immensely appreciate any tidy-up suggestions and verifications.