## Jonask Group Title find all pairs of positive intergers (x,y) for 2 years ago 2 years ago

a)$3^x=2^xy+1$ b)$x^y=y^{x-y}$

2. experimentX

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

3. experimentX

x=1, y=1 also works where you will get 3 = 3

so we just try out numbers

5. experimentX

no ...we can't do that. diophantine equations have always been my weaknesses. I'm just fooling around.

6. experimentX

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

9. experimentX

for these types of question, you should probably ask mukushla till then let me fool around.

lol thanks a lot

11. experimentX

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12. experimentX

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13. experimentX

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.

14. mukushla

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$$

15. mukushla

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

16. mukushla

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)

17. experimentX

thanks man!! you always amaze me!!

18. mukushla

np bro :)

19. experimentX

haha