{Number Theory} Smallest integer solution

[youngstudent16]

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.

 

[Dick]

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?

 

[youngstudent16]

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?

 

[Dick]

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.

 

[youngstudent16]

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 $96$

 

[Dick]

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?