8. Find all non-negative integer solutions $(x, y, z, w)$ of the equation $2^{x} \cdot 3^{y}-5^{x} \cdot 7^{w}=1$.
8. Solution: (1) If $y=0$, then $2^{x}-5^{z} \cdot 7^{w}=1$. When $z \neq 0$, $2^{x} \equiv 1(\bmod 5)$, since $2^{4} \equiv 1(\bmod 5)$,
$\therefore 4 \mid x$, thus $3 \mid 2^{x}-1=5^{z} \cdot 7^{w}$, which is impossible,
$\therefore z=0,2^{x}-7^{w}=1$,
When $x=1,2,3$, $(x, w)=(1,0),(3,1)$;
When $x \geqslant 4$, $7^{w}=2^{x}-1 \equiv-1(\bmod 16)$, but when $w$ is even, $7^{w} \equiv 1(\bmod 16)$, when $w$ is odd, $7^{w} \equiv 7(\bmod 16)$, a contradiction. $\therefore$ The solutions in this case are $(1,0,0,0),(3,0,0,1)$;
(2) If $y>0$. (1) When $x=1$, $2 \cdot 3^{y}-5^{x} \cdot 7^{w}=1$, then $5^{x} \cdot 7^{w} \equiv-1(\bmod 3)$,
i.e., $(-1)^{z} \equiv-1(\bmod 3)$, $\therefore z$ is odd,
$\therefore 2 \cdot 3^{y} \equiv 1(\bmod 5)$, since $3^{4} \equiv 1(\bmod 5)$,
$\therefore y \equiv 1(\bmod 4)$, i.e., $y=4 m+1$.
When $w \neq 0$, $2 \cdot 3^{y} \equiv 1(\bmod 7)$, since $3^{6} \equiv 1(\bmod 7)$, $\therefore y \equiv 4(\bmod 6)$, $y=6 n+4=4 m+1$, i.e.,
$6 n=4 m-3$ a contradiction, $\therefore$ there must be $w=0,2 \cdot 3^{y}-5^{x}=1$,

When $y=1$, $z=1$; when $y \geqslant 2$, $5^{x} \equiv-1(\bmod 9)$, since $5^{6} \equiv 1(\bmod 9)$, $\therefore z \equiv 3(\bmod 6)$,
$5^{3}+1\left|5^{x}+1,7\right| 5^{x}+1=2 \cdot 3^{y}$ a contradiction!
$\therefore$ The solutions in this case are $(1,1,1,0)$,
(2) When $x \geqslant 2$, since $5^{z} \cdot 7^{w} \equiv-1(\bmod 4)$, $5^{x}$
$\cdot 7^{w} \equiv-1(\bmod 3)$, i.e., $(-1)^{w} \equiv-1(\bmod 4)$, $(-1)^{z} \equiv-1(\bmod 3)$, $\therefore w, z$ are both odd, $2^{x} \cdot 3^{y}$ $=5^{z} \cdot 7^{w}+1 \equiv 35+1 \equiv 4(\bmod 8)$,
$\therefore x=2, y=2 k$, the equation is $4 \cdot 3^{y}-5^{x} \cdot 7^{w}=$ $1$, $\therefore 4 \cdot 3^{y} \equiv 1(\bmod 5)$, $y \equiv 2(\bmod 4)$, $4 \cdot 3^{y} \equiv$ $1(\bmod 7)$, $y \equiv 2(\bmod 6)$, $\therefore y \equiv 2(\bmod 12)$,
Let $y=12 m+2(m \geqslant 0)$, then $5^{z} \cdot 7^{w}=4 \cdot$
$$
\begin{array}{l}
3^{12 m+2}-1=\left(2 \cdot 3^{6 m+1}-1\right)\left(2 \cdot 3^{6 m+1}+1\right), \\
\because 2 \cdot 3^{6 m+1}+1 \equiv 0(\bmod 7),\left(2 \cdot 3^{6 m+1}-1,\right. \\
\left.2 \cdot 3^{6 m+1}+1\right)=1, \therefore 5 \mid 2 \cdot 3^{6 m+1}-1, \\
\quad \therefore\left\{\begin{array}{l}
2 \cdot 3^{6 m+1}-1=5^{x}, \\
2 \cdot 3^{6 m+1}+1=7^{w},
\end{array} \text { if } m \geqslant 1, \text { then } 5^{z} \equiv\right.
\end{array}
$$
$-1(\bmod 9)$, by the proof in (1) it is impossible.
If $m=0$, then $y=2,\left\{\begin{array}{l}2 \cdot 3-1=5^{z} \\ 2 \cdot 3+1=7^{w}\end{array}\right.$, we get $z=w$ $=1$, we get the solution $(2,2,1,1)$. Therefore, all solutions to the original equation are: $(x, y, z, w)=(1,0,0,0),(3,0,0,1),(1,1,1$, $0),(2,2,1,1)$.