Question

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.

Answer 1

@Empty @ganeshie8

Answer 2

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.

Answer 3

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

Answer 4

Nope lol

Answer 5

Oh just got it!

Answer 6

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

Answer 7

well we need it to be positive

Answer 8

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

Answer 9

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

Answer 10

Yup, that's it

Answer 11

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

Answer 12

Yes, I understood it as well haha

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Actually can you just explain why x = y+z

Answer 17

Oh nvm I see it! yz = 2015

Answer 18

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

Answer 19

Yup haha I see now xD

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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 ?