## Astrophysics one year ago A question I found on a forum (number theory) http://puu.sh/jDdI7/3477d12a27.png I would think you would just solve for x, although I don't know number theory I would still like to see how it's done hehe.

1. Astrophysics

@Empty @ganeshie8

2. Astrophysics

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

3. freckles

do you know the answer? I just want to make a guess if you do... 2016 could be an x value not sure if it is the smallest or whatever

4. Astrophysics

Nope lol

5. Astrophysics

Oh just got it!

6. Empty

could -2016 be the smallest answer, I am sorta looking at it from this aspect while trying to keep us in the reals @freckles

7. freckles

well we need it to be positive

8. Empty

Well I squared both sides and from there I equated yz=2015 then factored with WA, so x=y+z

9. Empty

So to minimize x I picked the closest in value factors, x= 31+65 = 96

10. Astrophysics

Yup, that's it

11. Empty

That wasn't as bad as I thought it was gonna be

12. Astrophysics

Yes, I understood it as well haha

13. freckles

$\sqrt{x+2 \sqrt{2015}}=\sqrt{y}+\sqrt{z} \\ x+2 \sqrt{2015}=y+z+2 \sqrt{y} \sqrt{z} \\ x=y+z \\ 2 \sqrt{2015}=2 \sqrt{y} \sqrt{z} \\ \text{ so we have } \\ \sqrt{2015}=\sqrt{yz} \\ 2015=yz \\ y+z=x \\ \text{ and then } z=\frac{2015}{y} \\ x= y+\frac{2015}{y}$ $x'=1-\frac{2(2015)}{y^2} \\ 1=\frac{2(2015)}{y^2} \\ y^2=2(2015) \\ y= \pm \sqrt{2(2015)} \\ \text{ nearest positive integer is } y=63 \\ z=\frac{2015}{63} \approx 32 \\ x=y+z=63+32=95$ and I guess I chose the wrong roundings

14. Astrophysics

Ha, that's very clever way! Nice one @freckles

15. freckles

http://www.wolframalpha.com/input/?i=sqrt%2831%29%2Bsqrt%2865%29%3Dsqrt%2896%2B2sqrt%282015%29%29 empty's values are totally true my equation would be approximately true

16. Astrophysics

Actually can you just explain why x = y+z

17. Astrophysics

Oh nvm I see it! yz = 2015

18. freckles

compared both sides of the equation of $x+2 \sqrt{2015}=y+z+2 \sqrt{yz}$

19. Astrophysics

Yup haha I see now xD

20. Empty

I'm curious now about when will the gcd(x,yz)=1 ? Always?

21. Empty

Like if we changed the value of 2015 to something else, not a square.

22. Empty

I guess to be more clearer, gcd(x+y, x*y)=1

23. Empty

Ok ok what I'm saying is the spirit of the problem to me is this: Take any composite number that isn't a perfect square. Which two factors of n that satisfy x*y=n also minimize this expression: x+y ?