{Number Theory} Smallest integer solution

Click For Summary

Homework Help Overview

The problem involves finding the smallest positive integer solution for \(x\) in the equation \(\sqrt{x+2\sqrt{2015}}=\sqrt{y}+\sqrt{z}\), where \(y\) and \(z\) are also positive integers. The context is rooted in number theory, particularly in exploring relationships between integers and irrational numbers.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the algebraic manipulation of the equation to isolate \(x\) and explore the implications of irrational components in the expression. Questions arise about how to handle the irrational term \(2\sqrt{2015}\) and what conditions must be met for \(x\) to remain a positive integer.

Discussion Status

The discussion has seen participants offering hints and guidance on how to approach the problem, particularly regarding the cancellation of irrational terms. Some participants have reached a conclusion about the value of \(x\), while others emphasize the need for a more rigorous justification of the conditions under which the integers \(y\) and \(z\) must satisfy.

Contextual Notes

Participants note that \(x\), \(y\), and \(z\) must be positive integers, and there is an ongoing exploration of the implications of \(2015yz\) being a perfect square, which is suggested as a necessary condition for the solution.

youngstudent16
Messages
59
Reaction score
1

Homework Statement


Let ##x,y,z## be positive integers such that ##\sqrt{x+2\sqrt{2015}}=\sqrt{y}+\sqrt{z}## find the smallest possible value of ##x##

Homework Equations


Not even sure what to ask I'm trying to learn number theory doing problems and look up information by doing the problems. [/B]

The Attempt at a Solution



The only thing I have done so far is get it set equal to ##x## so that I could make do some algebra and the answer would pop up.
##x=2\sqrt{yz}+y+z-2\sqrt{2015}##

Thanks for any help.
 
Physics news on Phys.org
youngstudent16 said:

Homework Statement


Let ##x,y,z## be positive integers such that ##\sqrt{x+2\sqrt{2015}}=\sqrt{y}+\sqrt{z}## find the smallest possible value of ##x##

Homework Equations


Not even sure what to ask I'm trying to learn number theory doing problems and look up information by doing the problems. [/B]

The Attempt at a Solution



The only thing I have done so far is get it set equal to ##x## so that I could make do some algebra and the answer would pop up.
##x=2\sqrt{yz}+y+z-2\sqrt{2015}##

Thanks for any help.

Should be kind of clear where to go from there. ##x,y,z## are integers. ##2\sqrt{2015}## is irrational. Something else irrational must cancel it. What must it be?
 
Dick said:
Should be kind of clear where to go from there. ##x,y,z## are integers. ##2\sqrt{2015}## is irrational. Something else irrational must cancel it. What must it be?
But if it cancels it out wouldn't that make ##x=0## which I can't have?
 
youngstudent16 said:
But if it cancels it out wouldn't that make ##x=0## which I can't have?

There are only two things with can be irrational in your expression. What the other one? Equate them.
 
  • Like
Likes   Reactions: youngstudent16
Dick said:
There are only two things with can be irrational in your expression. What the other one? Equate them.
Thank you I'm little tired so I just didn't understand what you said the first time even though it was clear. I got the correct answer now with that hint which is ##96##
 
youngstudent16 said:
Thank you I'm little tired so I just didn't understand what you said the first time even though it was clear. I got the correct answer now with that hint which is ##96##

You're welcome. But my argument why is pretty dodgy. If you want to be more rigorous, can you show that ##2015yz## must be a perfect square? What would that imply?
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K