MHB Find Integer Pairs $x,y$ with Infinitely Many Solutions

[anemone]

Find all pairs of integers $x,\,y>3$ such that there exist infinitely many positive integers $k$ for which $\dfrac{k^x+k-1}{k^y+k^2-1}$ is an integer.

 

[anemone]

Solution of other:

Spoiler

The problem is saying find all pairs of integers $x,\,y>3$ for which $k^y+k^2-1$ is a factor of the polynomial $k^x+k-1$.

It's clear that $x>y$. Let $x=y+a$. We have

$k^x+k-1=k^a(k^y+k^2-1)+(1-k)(k^{a+1}+k^a-1)$

So $k^y+k^2-1$ divides $k^{a+1}+k^a-1$. Now, $k^y+k^2-1$ has a real root $\alpha\in(0,\,1)$, $\alpha$ is also a root of $k^{a+1}+k^a-1$. Thus, $\alpha^{a+1}+\alpha^a=1$ and $\alpha^{y}+\alpha^2=1$. But $a+1>y>3$ and so $\alpha^{y}+\alpha^2>\alpha^{a+1}+\alpha^a$ with equality iff $a+1=y$ and $a=2$. Thus, $(x,\,y)=(5,\,3)$ is the only possible solution.

 

[Albert1]

Solution of other:

Spoiler

The problem is saying find all pairs of integers $x,\,y>3$ for which $k^y+k^2-1$ is a factor of the polynomial $k^x+k-1$.

It's clear that $x>y$. Let $x=y+a$. We have

$k^x+k-1=k^a(k^y+k^2-1)+(1-k)(k^{a+1}+k^a-1)$

So $k^y+k^2-1$ divides $k^{a+1}+k^a-1$. Now, $k^y+k^2-1$ has a real root $\alpha\in(0,\,1)$, $\alpha$ is also a root of $k^{a+1}+k^a-1$. Thus, $\alpha^{a+1}+\alpha^a=1$ and $\alpha^{y}+\alpha^2=1$. But $a+1>y>3$ and so $\alpha^{y}+\alpha^2>\alpha^{a+1}+\alpha^a$ with equality iff $a+1=y$ and $a=2$. Thus, $(x,\,y)=(5,\,3)$ is the only possible solution.

the pairs of x,y>3 are given
but your solution: (x,y)=(5,3)

 

[anemone]

the pairs of x,y>3 are given
but your solution: (x,y)=(5,3)

Ops! Seems to me this problem is set for the loose inequality $x,\,y\ge 3$, thank you very much Albert for pointing this out! I appreciate it!(Nod)

Sorry for the late reply too...:(