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

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In summary, having infinitely many solutions for $x$ and $y$ means that there is no limit to the number of pairs of integers that can be substituted into the equation to make it true. One way to find integer pairs with infinitely many solutions is to look for patterns in the equation. It is possible to prove that there are infinitely many solutions by showing that there is no limit to the number of possible solutions. There may be restrictions on the values of $x$ and $y$ for there to be infinitely many solutions, and there are real-world applications for finding such pairs in physics, engineering, and mathematics.
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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.
 
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  • #2
Solution of other:

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.
 
  • #3
anemone said:
Solution of other:

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)
 
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  • #4
Albert said:
the pairs of x,y>3 are given
but your solution: (x,y)=(5,3)

Ops!:eek: 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...:(
 
  • #5


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
 

FAQ: Find Integer Pairs $x,y$ with Infinitely Many Solutions

What does it mean to have infinitely many solutions for $x$ and $y$?

Having infinitely many solutions for $x$ and $y$ means that there are an infinite number of pairs of integers that can be substituted into the equation to make it true. This means that there is no limit to the number of possible solutions.

How can I find integer pairs $x$ and $y$ with infinitely many solutions?

One way to find integer pairs $x$ and $y$ with infinitely many solutions is to look for patterns in the equation. For example, if the equation involves squared terms or absolute values, there may be an infinite number of solutions. Additionally, you can try systematically plugging in different integer values for $x$ and solving for $y$ to see if there is a pattern.

Can I prove that there are infinitely many solutions for $x$ and $y$?

Yes, you can prove that there are infinitely many solutions for $x$ and $y$ by showing that there is no limit to the number of possible solutions. This can be done by finding a general formula or pattern for the solutions, or by showing that the equation can be rewritten in a form that clearly has an infinite number of solutions.

Are there any restrictions on the values of $x$ and $y$ for there to be infinitely many solutions?

Yes, there may be restrictions on the values of $x$ and $y$ in order for there to be infinitely many solutions. For example, the equation may only have infinitely many solutions for certain values of $x$ or $y$, or there may be a range of values for $x$ and $y$ that will result in infinitely many solutions.

Are there any real-world applications for finding integer pairs $x$ and $y$ with infinitely many solutions?

Yes, there are many real-world applications for finding integer pairs $x$ and $y$ with infinitely many solutions. For example, in physics and engineering, equations with infinitely many solutions can represent physical systems that have an infinite number of possible states. In mathematics, finding integer pairs with infinitely many solutions can help in proving conjectures or solving complex problems.

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