Hi, I've to solve the following diophantine equation (in positive integers):

$\displaystyle 3^x-5^y=z^2$.

Thanks for any help.

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- Jul 6th 2010, 08:01 PMArczi1984Diophantine equation
Hi, I've to solve the following diophantine equation (in positive integers):

$\displaystyle 3^x-5^y=z^2$.

Thanks for any help. - Jul 6th 2010, 08:34 PMAlso sprach Zarathustra
Maybe if we look the diophantine equation above $\displaystyle \mathbb{Z}_4=\{0,1,2,3\}$.

It seems (only in theory!) that the above diophantine equation had no solution in positive integers...

http://i48.tinypic.com/vopixx.jpg - Jul 7th 2010, 12:18 AMOpalg
- Jul 7th 2010, 12:27 AMsimplependulum
The solution is $\displaystyle (2,1,2) $

Take modulo $\displaystyle 4 $ we have

$\displaystyle (-1)^x - 1 \equiv z^2 \bmod{4} $

$\displaystyle x = 2t $ otherwise if $\displaystyle x $ is odd we have $\displaystyle z^2 \equiv -2 \equiv 2 \bmod{4} $ which is impossible .

Thus we have $\displaystyle 3^{2t} - 5^y = z^2 $

$\displaystyle (3^t)^2 - z^2 = 5^y $

$\displaystyle (3^t - z )(3^t + z ) = 5^y $ which leads to

$\displaystyle 3^t - z = 5^{\alpha} ~,~ 3^t + z = 5^{\beta} $ where $\displaystyle \alpha + \beta = y $

$\displaystyle 3^t = \frac{ 5^{\beta} + 5^{\alpha} }{2} $ which is the multiple of $\displaystyle 5 $ if $\displaystyle \alpha \beta > 0 $ which is again impossible .

Thus $\displaystyle \alpha = 0 $ ( not $\displaystyle \beta $ because $\displaystyle \alpha \leq \beta $ ) but they can't be both zero as we are looking for positive integers $\displaystyle y $

$\displaystyle 3^t = \frac{ 1 + 5^y }{2} ~~~~,~ \beta = y-0 = y $

We now show that $\displaystyle (t,y) = (1,1) $ is the only solution .

Suppose $\displaystyle t \geq 2 $ so we have

$\displaystyle \frac{ 1 + 5^y }{2} \equiv 0 \bmod{9} ~ \implies 1+ 5^y \equiv 0 \bmod{9} $

If we write down the first six powers of $\displaystyle 5 $ modulo $\displaystyle 9 $ , we obtain : $\displaystyle 1,5,7,8,4,2,(1) $

we have exactly $\displaystyle 5^6 \equiv 1 \bmod{9}$ and exactly $\displaystyle 5^{3} \equiv -1 \bmod{9} $

We conclude that $\displaystyle y = 6k + 3 = 3(2k+1)$ for some non-negative integers $\displaystyle k $

Back to this equation : $\displaystyle 3^t = \frac{ 1 + 5^y }{2}$

write $\displaystyle y = 3(2k+1) = 3m $ we have

$\displaystyle 3^t = \frac{ (1+5^3)(1-5^3 + 5^6 - ... + 5^{3m-3} )}{2} = 9 \cdot 7P $ the multiple of $\displaystyle 7 $ which is also impossible .

Therefore , $\displaystyle 0 < t \leq 1 ~ \implies t = 1 $

$\displaystyle t=1 $ is the only hope , luckily , we obtain $\displaystyle y = 1 $ and $\displaystyle z =2 $ , $\displaystyle (2,1,2) $ is the only solution . - Jul 7th 2010, 06:18 AMchiph588@