see here

Assume $Q$ is the polynomial with the smallest degree satisfied the statement Claim 1: $deg(Q) = p-2$ Proof: $\bullet$ $deq(Q)>p-1$. Then use the identity $x^p \equiv x \textrm{(mod p)}$, we get a contradiction with $deg(Q)$ is smallest $\bullet$ $deq(Q)=p-1$. Then we can use the identity $x^{p-1} \equiv 1 \textrm{(mod p)}$ because the problem is just care about $\overline{1,p-1}$ $\bullet$ $deg(Q)<p-2$. We have two case to kill: Case 1: $(Q(i),p)=1, \forall i=\overline{1,p-1}$ We have $\left\{Q(1),Q(2),..,Q(p-1) \right\}$ and $\left\{ 1\times Q(1),2\times Q(2),...,(p-1)\times Q(p-1)\right\}$ are two full residues mod p So by Wilson theorem, we get: $\left\{\begin{matrix}Q(1)\times Q(2)\times ...\times Q(p-1)\equiv -1 (\textrm{mod p}) & \\ & \\ (1.Q(1))\times (2.Q(2))... \times ((p-1).Q(p-1))\equiv - 1 (\textrm{mod p}) \end{matrix}\right.$ But $$(1.Q(1))\times (2.Q(2))...\times ((p-1).Q(p-1)) \equiv (1.2...(p-1))\times \left [ Q(1).Q(2)...Q(p-1) \right ]\equiv (-1).(-1) \equiv 1 (\textrm{mod p})$$a contradiction Case 2: Exist $i: Q(i) \vdots p$ Because $deg(Q)<p-2$, $deg(xQ(x)) \leq p-2$ . So we have the identity: $\sum_{i=0}^{p-1}iQ(i)\equiv 0 (\textrm{mod p}) \leftrightarrow \sum_{i=1}^{p-1}iQ(i)\equiv 0 (\textrm{mod p}), (1)$ But exist $i:Q(i) \vdots p$ and $\left\{ 1\times Q(1),2\times Q(2),...,(i-1)Q(i-1),(i+1)Q(i+1),...(p-1)\times Q(p-1)\right\} \sim (1,2,...x-1,x+1,...,p-1)$ From $(1)$, we get: $1+2+...+(p-1)-x \equiv 0 (\textrm{mod p}) \leftrightarrow x\equiv 0 (\textrm{mod p}) $, a contradiction Q.E.D

I haven't got any ideas, can someone help me ? one years back, is it just $Q(x)=x^{p-2}+1$ ?