$$
17 x^{19}-4 x^{17}-17 x^{15}+4=0
$$

all positive real solutions.
The equation to be solved is
$$
17\left(x^{19}-x^{15}\right)=4\left(x^{17}-1\right) \text {. }
$$

Obviously, $x=1$ is a solution.
When $x-1 \neq 0$, we can divide both sides by the factor $x-1$ to get
$$
\begin{array}{l}
17 x^{15}\left(x^{3}+x^{2}+x+1\right)=4\left(x^{16}+\cdots+x+1\right) \\
\Rightarrow \frac{x^{18}+x^{17}+x^{16}+x^{15}}{4}=\frac{x^{16}+\cdots+x+1}{17} .
\end{array}
$$

If $x>1$, then $x^{18}>x^{17}>\cdots>x>1$.
Thus, the left side of equation (1) is the average of the largest 4 numbers, while the right side is the average of the smallest 17 numbers, indicating that the left side $>$ the right side, meaning that $x>1$ is not a solution to the equation.
If $x<1$, similarly, it is also not a solution.
In summary, $x=1$ is the only positive real solution.