<s>[INST] Find all $x,y,z \in \mathbb{Z^{+}}$ such that $20^x+15^y=2015^z$ Find all $x,y,z \in \mathbb{Z^{+}}$ such that $20^x+15^y=2015^z$
I was checking modulo $4$ to see that $y,z$ must have the same parity.
Then took two cases when both $y,z$ are even  and when both are odd. But it is difficult to find solutions
 [/INST] $20^x+15^y=2015^z$ equiv. $4^x5^x+ 3^y 5^y=403^z 5^z$. 
We get $x=y$ or $x=z$ or $y=z$ because of divisibility by 5 to a certain power.
(Suppose, let's say, $x < y < z$ then $4^x =403^z 5^{z-x} - 3^y 5^{y-x}$ impossible because the left side is multiple of 5)
Case 1) $x=y$
This case $z \ge x$ (otherwise $(3^x + 4^x)5^{x-z}=403^z$, impossible because $403$ is not divisible by $5$)
and $3^x + 4^x=403^z5^{z-x} \tag1$
But $3^x + 4^x < 2\cdot4^x< 403^z$ so (1) cannot hold
Case 2) $y=z$
This case $x \ge y$ and $3^x5^{x-y} + 4^y=403^y \tag2$
From (2) $ 3^x5^{x-y}=403^y - 4^y=399 \cdot A = 3\cdot7\cdot19\cdot A \tag3$ 
So the equality (3) cannot hold.
Case 3) $x=z$
This case $y \ge x$ and $3^x + 4^y5^{y-x}=403^x \tag4$
Applying modulo $3$ to (4) we get $2^{y-x}=1 \mod 3$ so $y-x=2k, k \in \mathbb{N}$.
Also from (4) $ 4^y5^{y-x}=403^x - 3^x=400 \cdot A \tag5=4^25^2\cdot A$
So far, this case I couldn't get further.
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