
Problem 3. Solve in positive integers the equation:

$$
1+5^{x}=2^{y}+2^{z} \cdot 5^{t}
$$


Solution. If $y \geq 2$ and $z \geq 2$, then the right side of the given equation is divisible by 4 . But $1+5^{x} \equiv 2(\bmod 4)$ and hence $\min (y, z)=1$. On the other hand $2^{y} \equiv 1(\bmod 5)$. Thus $y$ is divisible by 4 ( 4 is the index of 2 modulo 5). It follows from here that $y \geq 4, z=\min (y, z)=1$ and the equation is $1+5^{x}=2^{4 y_{0}}+2.5^{t}$, where $y=4 y_{0}$ ( $y_{0}$ is a positive integer).

If $t=1$ then $5^{x}-2^{4 y 0}=9$. Using congruence modulo 3 we obtain that $x$ is even, i.e. $x=2 x_{0}$. Hence $\left(5^{x_{0}}-2^{2 y_{0}}\right)\left(5^{x_{0}}+2^{2 y_{0}}\right)=9$, from where $x_{0}=y_{0}=1$, i.e. $x=2$ and $y=4$.

Let $t>1$. Then $16^{y_{0}} \equiv 1(\bmod 25)$ and $y_{0}$ is divisible by $5(5$ is the index of 3 modulo 25). Hence $y_{0}=5 y_{1}$ and $1+5^{x}=2^{20 y_{1}}+2.5^{t}$. On the other hand $2^{10} \equiv 1(\bmod 11)$ and $5^{x} \equiv 2.5^{t}$

$(\bmod 11)$. Obviously $x>t$. Thus $5^{x-t} \equiv 2(\bmod 11)$, which is not true.

Therefore the given equation has a unique solution $(x, y, z, t)=(2,4,1,1)$.

## Grade 12 .
