Post Number:#2 by ahmedittihad » Thu Jun 22, 2017 5:42 am
The left hand side is even, so $ x^4+y^2$ is even, and $ x, y$ has the same parity.
However, if $ x, y$ are odd, we have that $ 4$ divides $ 7^x-3^y$, but $ 4$ doesn't divide $ x^4+y^2$.
So $ x, y$ are even, and let's say $ x=2x_1,y=2y_1$.
You'll get $ (7^{x_1}-3^{y_1})(7^{x_1}+3^{y_1})$ divides $ 4(4x_1^4+y_1^2)$
If $ y_1$ is odd, then $ 8$ divides $ (7^{x_1}-3^{y_1})(7^{x_1}+3^{y_1})$ but not $ 4(4x_1^4+y_1^2)$, so $ y_1$ is even.
Say $ y_1=2z_1$.
You'll get $ (7^{x_1}-3^{2z_1})(7^{x_1}+3^{2z_1})$ divides $ 16(x_1^4+z_1^2)$
So $ |(7^{x_1}-3^{2z_1})(7^{x_1}+3^{2z_1})|\leq |16(x_1^4+z_1^2)|$
But $ |(7^{x_1}-3^{2z_1})\geq 2$ (because both terms are even), and so $ 2(7^{x_1}+3^{2z_1})\leq 16(x_1^4+z_1^2)$.
But for $ x_1=1,2,3$ we easily get $ z_1=1,2$, and for $ x_1\geq 4$, we have that $ 2(7^{x_1})>16x_1^4$. Also, $ 2(3^{2z_1})>16z_1^2$ for all $ z_1\in \mathbb{N}$.
So it is enough to check $ (x_1,z_1)=(1,1),(2,1),(3,1),(1,2),(2,2),(3,2)$. We'll get $ (x_1,z_1)=1,1$, or $(x,y)=(2,4)$ as the only solution.
Frankly, my dear, I don't give a damn.