Problem 10.3. Find all positive integers $t, x, y, z$ such that

$$
2^{t}=3^{x} 5^{y}+7^{z}
$$
10.3. It follows that $2^{t} \equiv 1(\bmod 3)$ and therefore $t$ is even. Also $2^{t} \equiv 2^{z}$ $(\bmod 5)$, i.e. $2^{t-z} \equiv 1(\bmod 5)($ obviously $t>z)$. Then 4 divides $t-z$ and hence 2 divides $z$.

Further, it is clear that $t \geq 6>2$ and therefore $0 \equiv 3^{x}(-3)^{y}+(-1)^{z}$ $(\bmod 8)$ or, equivalently, $3^{x+y} \equiv(-1)^{y+1}(\bmod 8)$. If $y$ is even, then $3^{x+y} \equiv-1$ $(\bmod 8)$, a contradiction. Hence $y$ is odd and $3^{x+y} \equiv 1(\bmod 8)$.

It follows that $x+y$ is even and hence $x$ is odd. Set $t=2 m(m \geq 3), z=2 n$ ( $n \geq 1$ ) and write the equation in the form

$$
\left(2^{m}-7^{n}\right)\left(2^{m}+7^{n}\right)=3^{x} 5^{y}
$$

Since $\left(2^{m}-7^{n}, 2^{m}+7^{n}\right)=1$, the following three cases are possible:

Case 1. $2^{m}-7^{n}=3^{x}, 2^{m}+7^{n}=5^{y}$

Case 2. $2^{m}-7^{n}=5^{y}, 2^{m}+7^{n}=3^{x}$

Case 3. $2^{m}-7^{n}=1,2^{m}+7^{n}=3^{x} 5^{y}$.

In the first two cases we have $2^{m} \mp 7^{n}=3^{x}$. Having in mind that $m \geq 3$ and $x$ is odd, we get $\mp(-1)^{n} \equiv 3(\bmod 8)$, i.e. $3 \equiv \pm 1(\bmod 8)$, a contradiction.

In the third case the equality $2^{m}-7^{n}=1$ implies that $2^{m} \equiv 1(\bmod 7)$. Hence 3 divides $m$. Set $m=3 k$. It follows that $\left(2^{k}-1\right)\left(2^{2 k}+2^{k}+1\right)=7^{n}$. It is easy to see that $\left(2^{k}-1,2^{2 k}+2^{k}+1\right)$ equals 1 or 3 . Hence $2^{k}-1=1$, $2^{2 k}+2^{k}+1=7^{n}$. Then $k=1, n=1, m=3, t=6, z=2$ and we get $x=y=1$.

In conclusion, the only solution of the problem is $t=6, x=1, y=1$, $z=2$.