<s>[INST] if $xP(x)=x^t+P(x-1)$ find the $t\equiv ?\pmod 3$ 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
$$a_{t-2}x^{t-1}+a_{t-3}x^{t-2}+\cdots+a_{1}x^2+a_{0}x=(x-1)^{t-1}+a_{t-2}(x-1)^{t-2}+\cdots+a_{1}(x-1)+a_{0}$$
 [/INST] 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 $xP(x) = x^t+P(x-1)$.
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