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Jonask

find all pairs of positive intergers (x,y) for

  • one year ago
  • one year ago

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  1. Jonask
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    a)\[3^x=2^xy+1\] b)\[x^y=y^{x-y}\]

    • one year ago
  2. experimentX
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    3^x = 2^x y + 1 y =0, x=0 <-- this works ... since the lhs is power of 3

    • one year ago
  3. experimentX
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    x=1, y=1 also works where you will get 3 = 3

    • one year ago
  4. Jonask
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    so we just try out numbers

    • one year ago
  5. experimentX
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    no ...we can't do that. diophantine equations have always been my weaknesses. I'm just fooling around.

    • one year ago
  6. experimentX
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    if 3^n -1 is of form divisible by 2^n, then we have solutions of it.

    • one year ago
  7. Jonask
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    actually i never knew that they are called that diophantine

    • one year ago
  8. Jonask
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    \[3^x-2^xy=1\] but we can try x=1,2,3,4 and see if they yield the y value to give 1

    • one year ago
  9. experimentX
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    for these types of question, you should probably ask mukushla till then let me fool around.

    • one year ago
  10. Jonask
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    lol thanks a lot

    • one year ago
  11. experimentX
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    |dw:1350060934777:dw|

    • one year ago
  12. experimentX
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    |dw:1350061042671:dw|

    • one year ago
  13. experimentX
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    If, every C(n,i) is divisible by 2^(n-i) then we have integer solutions. we can check this manually ... since for n>N, this will be invalid.

    • one year ago
  14. mukushla
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    for the first one x=1 is answer for x>1 conclude that x must be even from\[3^{x-1}+3^{x-2}+...+3+1=2^{x-1}y\]and setting x=2k going back to original equation\[3^{2k}-1=2^{2k}y\]\[(3^k-1)(3^k+1)=2^{2k}y\]now since \(\gcd(3^k-1,3^k+1)=2\) one of \(3^k-1 \ \ , \ 3^k+1\) must be divisible by \(2^{2k-1}\) and this is possible onle when \(k<3\)

    • one year ago
  15. mukushla
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    for the second one note that\[(\frac{x}{y})^y=y^{x-2y}\] so \(x\ge y\) and \(y|x\) so set \[x=ky\]and find the limits for k

    • one year ago
  16. mukushla
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    if u wanna check the solutions are for the first one (x,y)=(1,1),(2,2),(4,5) for the second one (x,y)=(1,1),(9,3),(8,2)

    • one year ago
  17. experimentX
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    thanks man!! you always amaze me!!

    • one year ago
  18. mukushla
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    np bro :)

    • one year ago
  19. experimentX
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    haha

    • one year ago
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