<s>[INST] How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers. Problem: How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers.

I tried a similar approach to the ones described here, but to no avail. I can't split up the RHS so that all the factors have the same base. 
 [/INST] The product $n=3^8\cdot 5^6\cdot13^4$ has a total of $(8+1)(6+1)(4+1)=315$ factors $d$. Those can be paired up as $(d,n/d)$. For the choice $d=n/d=\sqrt n$ the two factors are equal. For the remaining $157=(315-1)/2$ pairs we must use the bigger one as $x+y$ and the smaller as $x-y$. Because all the factors $d, n/d$ are odd, the resulting system has a solution $(x,y)$ in integers.
The answer is thus $157$. They come from solutions of the system $x+y=n/d, x-y=d$, that is,
$$
x=\frac{(n/d)+d}2,\quad y=\frac{(n/d)-d}2,
$$
with $d$ ranging over the set of factors $<\sqrt n$.
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