Find $p,q $ prime numbers s.t. $p+p^2+...+p^{10}-q=2017$ 
Find $p,q $ prime numbers s.t. $$p+p^2+\cdots+p^{10}-q=2017.$$

It's easy to see that $p=2; q=29$ is solution.
There exists another solutions?

Partial answer:
LHS is odd only when $q$ is odd. Consider the two cases:


*

*$p$ is even

*$p$ is odd
For the first case, $p=2$, forcing $q=29$.
For the second case, $p=2a+1$ and $q=2b+1$, with $a,b\in\mathbb{N}$, so we solve $$\left(\sum_{i=1}^{10}p^i\right)-q=2017$$ giving $$\frac{p^{11}-p}{p-1}=2016-2b\implies (2a+1)^{11}-2a-1=2a(2016-2b)$$ or $$(2a+1)^{11}=2a(2017-2b)+1\implies c=2017-2b$$ where $$\begin{align}c&=1024a^{10}+5632a^9+14080a^8+21120a^7+21120a^6+14784a^5+7392a^4+2640a^3+660a^2+110a+11\\&=2d+11\end{align}$$ This means that $$b+d=1003$$ where $$\begin{align}d&=a(512a^9+2816a^8+7040a^7+10560a^6+10560a^5+7392a^4+3696a^3+21320a^2+330a+55)\end{align}$$
