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Jonask
 3 years ago
find all pairs of positive intergers (x,y) for
Jonask
 3 years ago
find all pairs of positive intergers (x,y) for

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Jonask
 3 years ago
Best ResponseYou've already chosen the best response.0a)\[3^x=2^xy+1\] b)\[x^y=y^{xy}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.33^x = 2^x y + 1 y =0, x=0 < this works ... since the lhs is power of 3

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3x=1, y=1 also works where you will get 3 = 3

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.0so we just try out numbers

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3no ...we can't do that. diophantine equations have always been my weaknesses. I'm just fooling around.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3if 3^n 1 is of form divisible by 2^n, then we have solutions of it.

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.0actually i never knew that they are called that diophantine

Jonask
 3 years ago
Best ResponseYou've already chosen the best response.0\[3^x2^xy=1\] but we can try x=1,2,3,4 and see if they yield the y value to give 1

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3for these types of question, you should probably ask mukushla till then let me fool around.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350060934777:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3dw:1350061042671:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3If, every C(n,i) is divisible by 2^(ni) then we have integer solutions. we can check this manually ... since for n>N, this will be invalid.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1for the first one x=1 is answer for x>1 conclude that x must be even from\[3^{x1}+3^{x2}+...+3+1=2^{x1}y\]and setting x=2k going back to original equation\[3^{2k}1=2^{2k}y\]\[(3^k1)(3^k+1)=2^{2k}y\]now since \(\gcd(3^k1,3^k+1)=2\) one of \(3^k1 \ \ , \ 3^k+1\) must be divisible by \(2^{2k1}\) and this is possible onle when \(k<3\)

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1for the second one note that\[(\frac{x}{y})^y=y^{x2y}\] so \(x\ge y\) and \(yx\) so set \[x=ky\]and find the limits for k

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.1if 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)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.3thanks man!! you always amaze me!!
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