9. Find all positive integer solutions $(x, y, z, t)$ to the equation $5^{x}+1=2^{y}+2^{x}-5^{t}$.

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9. Solution: Let $(x, y, z, t)$ be a positive integer solution of the equation, then $2^{y} \equiv 1(\bmod 5), \because 2^{4} \equiv 1(\bmod 5), \therefore 4 \mid y$,
Taking both sides of the equation $\bmod 4$ we get $2^{x} \equiv 2(\bmod 4)$,
$$
\therefore z=1 \text {. }
$$

Let $y=4 r$, then $5^{x}+1=2^{4 r}+2 \cdot 5^{t}$, i.e., $5^{x}-2=16^{r}-1$, the solution is $(x, y, z, t)=(2,4,1,1)$;
- $5^{t}=16^{r}-1$, taking both sides $\bmod 3$ we get:
$(-1)^{x}+(-1)^{t} \equiv 0(\bmod 3), \therefore x, t$ are one odd and one even, $\because 5^{t} \equiv 1$ or $5(\bmod 8), \therefore$ for $5^{x}=2 \cdot 5^{t}+$ . $16^{r}-1$
Taking both sides $\bmod 8$ we get: $5^{x} \equiv 2 \cdot 5^{t}-1 \equiv 1(\bmod 8)$,
$\therefore x$ is even, $t$ is odd.
(1) If $t=1$, then $5^{x}=16^{r}+9$, let $x=2 m$, then $\left(5^{m}-3\right)\left(5^{m}+3\right)=16^{r}$.
$$
\begin{array}{l}
\because\left(5^{m}-3,5^{m}+3\right)=\left(5^{m}-3,6\right)=2, \\
\therefore 5^{m}-3=2,5^{m}+3=2^{4 r-1}, \\
\therefore m=1,2^{4 r-1}=2^{3}, \therefore r=1, y=4, x=2,
\end{array}
$$
(2) If $t>1$, then $t \geqslant 3, x \geqslant 4,5^{3} \mid 5^{x}-2 \cdot 5^{t}$ $=16^{r}-1$,
$$
\because 16^{r}-1=(15+1)^{r}=15^{r}+\cdots+C_{r}^{2} \cdot 15^{2}
$$
$+\mathrm{C}_{r}^{1} \cdot 15, \therefore 5 \mid r$, let $r=5 k$, then $16^{5 k}-1 \equiv 5^{5 k}-$ $1 \equiv\left(5 \cdot 3^{2}\right)^{k}-1 \equiv 0(\bmod 11)$,
$\therefore 11 \mid 5^{x}-2 \cdot 5^{t}=5^{t}\left(5^{x-t}-2\right)$, i.e., $11 \mid 5^{x-t}$ -2 , but $5^{n} \equiv 1,3,4,5,9(\bmod 11)$, i.e., $11 \times 5^{x-t}-$ 2 , contradiction! Hence the original equation has a unique solution $(x, y, z, t)=(2$, $4,1,1)$.