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...:(

 

[CellCoree]

I would first analyze the given information and try to understand the problem at hand. From the given statement, it is clear that we are looking for integer pairs $x$ and $y$ that satisfy a certain condition, which involves the variable $k$. Additionally, we are given a constraint that $x$ and $y$ must be greater than 3, and that there must be infinitely many positive integers $k$ that satisfy the given condition.

To find these integer pairs, I would first start by trying to simplify the expression given to us. We can rewrite the expression as:

$\dfrac{k^x+k-1}{k^y+k^2-1} = \dfrac{k^x+k-1}{(k+1)(k^{y-1}-k^{y-2}+...+k+1)}$

From this, we can see that in order for the expression to be an integer, the numerator must be divisible by the denominator. This means that $k^x+k-1$ must be divisible by $(k+1)(k^{y-1}-k^{y-2}+...+k+1)$.

Next, I would try to find a general pattern for values of $x$ and $y$ that satisfy this condition. It may be helpful to start with small values of $x$ and $y$ and see if a pattern emerges. For example, when $x=4$ and $y=5$, we get:

$\dfrac{k^4+k-1}{k^5+k^2-1} = \dfrac{(k+1)(k^3-k^2+k-1)}{(k+1)(k^4-k^3+k^2-k+1)} = \dfrac{k^3-k^2+k-1}{k^4-k^3+k^2-k+1}$

We can see that for any positive integer value of $k$, the numerator and denominator will be equal, resulting in an integer value for the expression. This means that for any value of $k$, the expression will be an integer, and thus there are infinitely many positive integer solutions for $k$.

Therefore, we can conclude that for any integer pairs $x=4$ and $y=5$, the given condition is satisfied and there are infinitely many positive integer solutions for $k$. By extending this pattern, we