# Thread: Diophantine equation

1. ## Diophantine equation

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

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

Thanks for any help.

2. Originally Posted by Arczi1984
Hi, I've to solve the following diophantine equation (in positive integers):

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

Thanks for any help.
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...

3. Originally Posted by Also sprach Zarathustra
It seems (only in theory!) that the above diophantine equation had no solution in positive integers...
It has at least one solution, because $\displaystyle 3^2-5^1 = 2^2$.

4. Originally Posted by Arczi1984
Hi, I've to solve the following diophantine equation (in positive integers):

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

Thanks for any help.
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 .

5. Originally Posted by simplependulum
$\displaystyle t=1$ is the only hope , luckily , we obtain $\displaystyle y = 1$ and $\displaystyle z =2$ , $\displaystyle (2,1,2)$ is the only solution .
Just in case you didn't follow how simplependulum obtained $\displaystyle y=1$, if $\displaystyle y>1$ then $\displaystyle z^2=3^2-5^y<0$.