Let polynomials $P(x)\in Z[x]$,and such $\deg{(P(x))}=t-1$,and such for any real $x$ have $$xP(x)=x^t+P(x-1)$$
Find the $t\equiv ?\pmod 3$?
I try Let $P(x)=x^{t-1}+a_{t-2}x^{t-2}+\cdots+a_{1}x+a_{0}$ where $a_{i}\in Z$
so we have $$a_{t-2}x^{t-1}+a_{t-3}x^{t-2}+\cdots+a_{1}x^2+a_{0}x=(x-1)^{t-1}+a_{t-2}(x-1)^{t-2}+\cdots+a_{1}(x-1)+a_{0}$$
Let $P(x) = \displaystyle\sum_{k = 0}^{t-1}a_kx^k$ where $a_k \in \mathbb{Z}$. Also, for $r = 0,1,\ldots,5$ define $S_r = \displaystyle\sum_{\substack{k = 0\\k \equiv r \pmod{6}}}^{t-1}a_k$.
Note that for $x = e^{i\tfrac{\pi}{3}} = \tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i$ we have $x-1 = -\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i = e^{i\tfrac{2\pi}{3}}$.
By plugging $x = e^{i\tfrac{\pi}{3}}$ into the identity $xP(x) - P(x-1) = x^t$ and using the fact that $e^{i\tfrac{\pi}{3}k}$ and $e^{i\tfrac{2\pi}{3}k}$ are $6$-periodic, we get:
$$e^{i\tfrac{\pi}{3}}P\left(e^{i\tfrac{\pi}{3}}\right) - P\left(e^{i\tfrac{2\pi}{3}}\right) = e^{i\tfrac{\pi}{3}t}$$
$$\sum_{k = 0}^{t-1}a_ke^{i\tfrac{\pi}{3}(k+1)} - \sum_{k = 0}^{t-1}a_ke^{i\tfrac{2\pi}{3}k} = e^{i\tfrac{\pi}{3}t}$$
$$\sum_{k = 0}^{t-1}a_k\left(e^{i\tfrac{\pi}{3}(k+1)} - e^{i\tfrac{2\pi}{3}k}\right) = e^{i\tfrac{\pi}{3}t}$$
$$\left[-\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i\right]S_0 + \left[-\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}i\right]S_2 + \left[-\tfrac{3}{2}-\tfrac{\sqrt{3}}{2}i\right]S_3 + \left[1-\sqrt{3}i\right]S_4 + \left[\tfrac{3}{2}+\tfrac{\sqrt{3}}{2}i\right]S_5 = e^{i\tfrac{\pi}{3}t}$$
Equating real and imaginary parts gives:
$$\tfrac{1}{2}\left[-S_0-S_2-3S_3+2S_4+3S_5\right] = \cos\left(\tfrac{\pi}{3}t\right)$$
$$\tfrac{\sqrt{3}}{2}\left[S_0+S_2-S_3-2S_4+S_5\right] = \sin\left(\tfrac{\pi}{3}t\right)$$
Then, $2$ times the real part equation plus $\tfrac{2}{\sqrt{3}}$ times the imaginary part equation gives:
$$4(S_5-S_3) = 2\cos\left(\tfrac{\pi}{3}t\right)+\tfrac{2}{\sqrt{3}}\sin\left(\tfrac{\pi}{3}t\right)$$
Since the $a_k$'s are integers, the $S_r$'s are also integers. So, $4(S_5-S_3)$ must be a multiple of $4$. But, $$2\cos\left(\tfrac{\pi}{3}t\right)+\tfrac{2}{\sqrt{3}}\sin\left(\tfrac{\pi}{3}t\right) = \begin{cases}2 & \text{if} \ t \equiv 0 \ \text{or} \ 1 \pmod{6} \\ 0 & \text{if} \ t \equiv 2 \ \text{or} \ 5 \pmod{6} \\ -2 & \text{if} \ t \equiv 3 \ \text{or} \ 4\pmod{6}\end{cases}.$$
Therefore, $t \equiv 2 \ \text{or} \ 5 \pmod{6}$, and thus, $\boxed{t \equiv 2 \pmod{3}}$.
Note: It is easy to check that $t = 2$ and $P(x) = x+1$ is a solution. There may be other solutions for larger $t \equiv 2\pmod{3}$.
Further Analysis: By substituting the sum for $P(x)$ into $xP(x) = x^t+P(x-1)$, we have:
$$\sum_{k = 0}^{t-1}a_kx^{k+1} = x^t + \sum_{n = 0}^{t-1}a_n(x-1)^n$$ $$\sum_{k = 1}^{t}a_{k-1}x^k = x^t + \sum_{n = 0}^{t-1}a_n\left(\sum_{k = 0}^{n}\dbinom{n}{k}(-1)^{n-k}x^k\right)$$ $$a_{t-1}x^t + \sum_{k = 1}^{t-1}a_{k-1}x^k = x^t + \sum_{k = 0}^{t-1}\left(\sum_{n = k}^{t-1}a_n\dbinom{n}{k}(-1)^{n-k}\right)x^k$$
By equating corresponding coefficients, we get:
$$a_{t-1} = 1$$
$$a_{k-1} = \sum_{n = k}^{t-1}a_n\dbinom{n}{k}(-1)^{n-k} \ \text{for} \ k = 1,\ldots,t-1$$
$$0 = \sum_{n = 0}^{t-1}a_n(-1)^{n}$$
The first two equations uniqely determine the coefficients $a_k$ for $k = 0,\ldots,t-1$. Then, the polynomial $P(x) = \displaystyle\sum_{k = 0}^{t-1}a_kx^k$ satisfies $xP(x) = x^t+P(x-1)$ iff the third equation is satisfied.
I tested this recursion for $1 \le t \le 56$ using a computer program. It appears that $t = 2$ was the only value of $t$ in that range for which $P(x)$ satisfies $xP(x) = x^t+P(x-1)$.
Given the equation $\space xP(x)=x^t+P(x-1)$, where $\deg(P(x))=t-1$, I would first solve for a general form for $P(x)$.
$$P(x)=x^{t-1}+\frac{P(x-1)}{x}$$
Where the general form for $P(x)$ is
$$P(x)=a_{t-1}x^{t-1}+a_{t-2}x^{t-2}+\dots+a_1x+a_0 = \sum_{i=0}^{t-1}a_ix^i$$
Plugging in to our equation yields
$$\sum_{i=0}^{t-1}{a_ix^i}=x^{t-1}+\frac{1}{x}\sum_{i=0}^{t-1}{a_i(x-1)^{i}}$$
Using the binomial theorem,
$$\sum_{i=0}^{t-1}{a_ix^i}=x^{t-1}+\sum_{i=0}^{t-1}{a_i{{t-1}\choose{i}}x^{i-1}(-1)^{t-1-i}}$$
Matching coefficients,
$$a_{t-1}=1\\a_{t-2}={{t-1}\choose{t-1}}a_{t-1}=1\\a_{t-3}=-{{t-1}\choose{t-2}}a_{t-2}=1-t\\a_{t-4}={{t-1}\choose{t-3}}a_{t-3}=\frac{(t-1)(t-2)}{2}a_{t-3}=\frac{1}{2}(t-1)^2(t-2)\\\dots\\a_{t-n}=(-1)^{n}{{t-1}\choose{t-n}}a_{t-n+1}$$
I hope this helps!
