Find number of solutions of $x^n-y^n = 2^{500}$, $x,y,n \in \mathbb{N}$ 
Let $x,y,n\in \mathbb{Z_{>0}}, n>1.$ How many solutions exist for the equation $x^n-y^n=2^{500}$. 

My attempt: Clearly, if a solution exists then $x>y>0$. Consider $n=2$, then $(x-y)(x+y)=2^{500}$ , let $u=x-y$ and $v=x+y$, which gives us $uv=2^{500}$ and $x=\frac{v+u}{2}$ , $y=\frac{v-u}{2}$. Where $u$ and $v$ are positve even integers and $v>u$.  Let $u = 2^i$ and $v=2^j$, which implies $i+j=500$ and $0<i<j<500$. Hence there are $249$ solutions for the above equation when $n=2$. They are $$(x,y)=(2^{j-1}+2^{i-1},2^{j-1}-2^{i-1}), \text{for }  0<i<j<500 , i+j=500 $$
However the above method cannot be used for $n>2$, for $n=3$, I tried using the substitutions $u=x-y$ , $v=xy$ and $u=x-y$ , $v=x^2-xy+y^2$. How do we solve this for a general $n$ ?      

Maybe there are no solutions of the above equation for higher values of $n$. For $n=4$ , let $u=x^2-y^2$ and $v=x^2+y^2$, which gives us $uv=2^{500}$ and $x^2=\frac{v+u}{2}$ , $y^2=\frac{v-u}{2}$. Where $u$ and $v$ are positve even integers and $v>u$. Then as we have solved above $x^2=2^{i-1}\left(2^{j-i}+1\right)$ and $y^2=2^{i-1}\left(2^{j-i}-1\right)$. For $x$ and $y$ to be integers , $i$ should be odd and both $2^{j-i}+1$ and $2^{j-i}-1$ should be perfect squares, but this is not possible and hence there are no solutions for 
$n=4$. Is this argument correct ? 
 A: Your argument is correct for handling the cases where $n = 2$ and $n = 4$. Next, as Ross Millikan's question comment states, Fermat' last theorem shows $n$ can't be a multiple of $4$. Also, your result shows this since, if $n = 4m$ for some positive integer $m$, you have
$$(x^m)^4 - (y^m)^4 = 2^{500} \tag{1}\label{eq1A}$$
so if you let $x_1 = x^m$ and $y_1 = y^m$, you get
$$(x_1)^4 - (y_1)^4 = 2^{500} \tag{2}\label{eq2A}$$
which is in the same form as your original equation and that you've proved has no solutions.
Next, consider $n = 2k$ where $k \gt 1$ is an odd integer. Your equation then becomes
$$(x^k)^2 - (y^k)^2 = 2^{500} \tag{3}\label{eq3A}$$
As you've shown, this means the solutions are, for some integers $0 \lt i \lt j \lt 500$, that
$$x^k = 2^{j-1} + 2^{i-1} = 2^{i-1}(2^{j-i} + 1) \tag{4}\label{eq4A}$$
$$y^k = 2^{j-1} - 2^{i-1} = 2^{i-1}(2^{j-i} - 1) \tag{5}\label{eq5A}$$
Note, any odd factors of $x$ or $y$ must be to the power of $k$ and be equal to $2^{j-1} + 1$ and $2^{j-i} - 1$ respectively. However, these $2$ values are only $2$ apart, but any integers to the power of $k$, for any odd $k \ge 3$, are always more than $2$ apart, even when they are consecutive, so this is not possible. Note that this actually applies to all $k \gt 1$, so it includes all even $n \gt 2$, such as the cases of $4$ and all larger multiples of $4$.
Finally, consider $n$ being an odd integer. You can factor $x - y$ out to get
$$(x-y)\left(\sum_{i=0}^{n-1} x^{n-1-i}y^{i}\right) = 2^{500} \tag{6}\label{eq6A}$$
Since the right side is a just a power of $2$, this means the only factors on the left can be $2$. As such, $x$ and $y$ must have the same parity, i.e., either both odd or both even. However, if they are both odd, then note the summation has $n$ positive terms, i.e., an odd number of them, so its sum is odd and, as $n \gt 1$, the sum is also $\gt 1$. This is not allowed, so both $x$ and $y$ must both be even, say $x = 2x_1$ and $y = 2y_1$. You can substitute this into the original equation and divide both sides by $2^{n}$ to get
$$(x_1)^n - (y_1)^n = 2^{500 - n} \tag{7}\label{eq7A}$$
You can then factor this again to get the same situation & issues as in \eqref{eq6A}, meaning both $x_1$ and $y_1$ are also even, so you can let $x_2 = 2x_1$ & $y_2 = 2y_1$ and then repeat dividing by $2^{n}$. You can keep doing this, say $s$ times, so the power of $2$ on the right is $500 - ns$, with this eventually being equal to $0$, i.e., a value of $1$, which is not allowed for $n \gt 1$, or it's less than $0$, which is also not possible. In either case, the result is there are no possible solutions for odd $n$.
Thus, in summary, the only possible solutions are the ones you found for $n = 2$.
