Why don't you try bashing this

The LHS is $2^{2015}>(2^{10})^{200}>10^{600}$ while the RHS is $<100*200*20^{100}<100*1000*100^{100}<10^{300}$ . Thus there are $\boxed{0}$ solutions.

We will investigate this by using Budan's Theorem.

Budan's Theorem states that the number of existing roots of a polynomial function $P(x)$ within a given interval $(a,b)$ cannot be greater than the difference in the number of sign variations in the coefficients of $P(x+a)$ and $P(x+b)$ . That is,

$\large n \leq s_{P(x+a)} - s_{P(x+b)}$

where $n$ is the number of possible roots in the given interval; $s$ is the number of sign variations in the coefficients of a given polynomial

This is all provided that the following conditions are met:

$P(x+a) \geq P(x+b)$ , and

$P(b) \neq 0$ .

So we check.

In both polynomials, that is, $P(x+17)$ and $P(x+18)$ , we are assured that there will be no negative coefficients occurring anywhere in the function where x is multiplied. Thus, there is only one sign variation which both occurs at the end of each function. That is, at the coefficient in x and the constant. So, $s_{P(x+a)} = s_{P(x+b)} = 1$

$\large n \leq s_{P(x+a)} - s_{P(x+b)}$

$\large n = 0$

Which means that there are no real roots in the said interval. :)

There are other means to solve this problem. I just wanted to use this theorem. Don't forget to reshare!