Find all solutions of $9^x-4^y=5^z$ over the naturals Find all solutions of $9^x-4^y=5^z$ over the naturals.
My attempt: We have
$${(3^x)}^2-{(2^y)}^2=5^z \Rightarrow (3^x+2^y)(3^x-2^y)=5^z.$$
It can be easily proved that $\gcd (3^x+2^y,3^x-2^y)=1$ so we have
$$3^x-2^y=1,3^x+2^y=5^z.$$
What do we have to do from here?

As @ Xam said,
if $y=0$, then there is no solution,
if $y=1$, then $x=1$.
Now if $y$ is greater than $1$, then $2^y$ is divisible by $4$.
using modular arithmetic module $4$ we have:
$(-1)^x-0$ is congruent to $1$ module $4$, so $x$ must be even, i.e. $x=2k$.


Now $2^y=3^{2k}-1=(3^k-1)(3^k+1)$, and by the same conclusion you have done, as you explained, we can conclude that:
$(3^k-1)=2$ and $(3^k+1)=2^{y-1}$, 
so $k=1$ and $x=2$ and $y-1=2$.
But this values for $x$ and $y$ does not give any solution to the main equation.




Another soloution:
We claim that $y$ could not be greater than $1$.
Suppose on contrary, that y is greater than $1$, then $4^y$ is divisible by $8$.
Using modular arithmetic module $8$ we have:
$1-0$ is congruent to $5^z$ module $8$, so $z$ must be even, i.e. $z=2t$.
Again by considering the main equation module $3$, we have :
$0-1$ is congruent to $(-1^2)^t$ module $3$, i.e. $-1=1$ module $3$, which is a contradiction.


So we have two cases:


*

*$y=1$, then $(3^x-2)(3^x+2)=5^z$, and by the same conclusion as you have been used, we can conclude that:
$(3^x-2)=1$ and $(3^x+2)=5^z$, which gives:
$x=1$ and $z=1$.

*$y=0$, then $(3^x-1)(3^x+1)=5^z$, we can conclude that:
$(3^x-1)=2$ and $(3^x+1)=(1/2)5^z$, which does'nt gives any solution.
