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BdMO Online Forum • View topic - Divisibility... with x,y

Divisibility... with x,y

For discussing Olympiad Level Number Theory problems

Divisibility... with x,y

Determine all pairs of positive integers $(x,y)$ such that
$(7^x-3^y)|(x^4+y^2)$
Katy729

Posts: 37
Joined: Sat May 06, 2017 2:30 am

Re: Divisibility... with x,y

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.

Posts: 147
Joined: Mon Mar 28, 2016 6:21 pm