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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.

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  1. Astrophysics
    • one year ago
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    @Empty @ganeshie8

  2. Astrophysics
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Nope lol

  5. Astrophysics
    • one year ago
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    Oh just got it!

  6. Empty
    • one year ago
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    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
    • one year ago
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    well we need it to be positive

  8. Empty
    • one year ago
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    Well I squared both sides and from there I equated yz=2015 then factored with WA, so x=y+z

  9. Empty
    • one year ago
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    So to minimize x I picked the closest in value factors, x= 31+65 = 96

  10. Astrophysics
    • one year ago
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    Yup, that's it

  11. Empty
    • one year ago
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    That wasn't as bad as I thought it was gonna be

  12. Astrophysics
    • one year ago
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    Yes, I understood it as well haha

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

  14. Astrophysics
    • one year ago
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    Ha, that's very clever way! Nice one @freckles

  15. freckles
    • one year ago
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    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
    • one year ago
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    Actually can you just explain why x = y+z

  17. Astrophysics
    • one year ago
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    Oh nvm I see it! yz = 2015

  18. freckles
    • one year ago
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    compared both sides of the equation of \[x+2 \sqrt{2015}=y+z+2 \sqrt{yz}\]

  19. Astrophysics
    • one year ago
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    Yup haha I see now xD

  20. Empty
    • one year ago
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    I'm curious now about when will the gcd(x,yz)=1 ? Always?

  21. Empty
    • one year ago
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    Like if we changed the value of 2015 to something else, not a square.

  22. Empty
    • one year ago
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    I guess to be more clearer, gcd(x+y, x*y)=1

  23. Empty
    • one year ago
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    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 ?

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