Is it possible to find an integer solution $r≥4$ to an equation?

Question

Is it possible to find an integer solution $r≥4$ to this equation?

$$11r²³-7r²¹+11r¹⁸-7r¹⁶-2r¹²+11r¹¹- 7r⁹-2r⁷-2 =0$$

I try some special values of $r$ but without any sucess.

Answer 1

Note that by moving the $2$ to the RHS we have

$$r^7(11r^{16}-7r^{14}+11r^{11}-7r^9-2r^5-7r^2-2)=2$$

if $r\in\Bbb Z$, then $r|2$ i.e. $rk=2$ for some integer $k$, in this case for

$$k=r^6(11r^{16}-7r^{14}+11r^{11}-7r^9-2r^5-7r^2-2).$$

But $2$ is a prime so $r\in\{-2,-1,1,2\}$, so there are no possible solutions bigger than $4$.

Answer 2

An integer solution $r$ of a polynomial $\sum\limits_{i=0}^n a_ix^i$ has to be a factor of $a_0$, i.e. $rk=a_0$ for an $k\in\mathbb{Z}$. So for your polynomial the only possible integer solution would be $1,-1,2,-2$.