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

Mathematics
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a)\[3^x=2^xy+1\] b)\[x^y=y^{x-y}\]
3^x = 2^x y + 1 y =0, x=0 <-- this works ... since the lhs is power of 3
x=1, y=1 also works where you will get 3 = 3

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Other answers:

so we just try out numbers
no ...we can't do that. diophantine equations have always been my weaknesses. I'm just fooling around.
if 3^n -1 is of form divisible by 2^n, then we have solutions of it.
actually i never knew that they are called that diophantine
\[3^x-2^xy=1\] but we can try x=1,2,3,4 and see if they yield the y value to give 1
for these types of question, you should probably ask mukushla till then let me fool around.
lol thanks a lot
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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.
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 \(k<3\)
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
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)
thanks man!! you always amaze me!!
np bro :)
haha

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