Six, find all non-negative integer solutions $(x, y, z, w)$ of the equation $2^{x} \cdot 3^{y}-5^{x} \cdot 7^{w}=1$.
(Supplied by Chen Yonggao)
Given that $5^{x} \cdot 7^{x}+1$ is even, we know $x \geqslant 1$.
Case 1 If $y=0$, then
$$
2^{x}-5^{x} \cdot 7^{x}=1 \text {. }
$$

If $z \neq 0$, then $2^{x} \equiv 1(\bmod 5)$, which implies $41 x$.
Therefore, $31\left(2^{x}-1\right)$, which contradicts $2^{x}-5^{x} \cdot 7^{x}=1$.
If $z=0$, then $2^{x}-7^{x}=1$.
For $x=1,2,3$, direct calculation yields
$$
(x, w)=(1,0),(3,1) \text {. }
$$

For $x \geqslant 4$, we have $7^{*} \equiv-1(\bmod 16)$, and direct calculation shows this is impossible.
Thus, when $y=0$, all non-negative integer solutions are
$$
(x, y, z, w)=(1,0,0,0),(3,0,0,1) \text {. }
$$

Case 2 If $y>0, x=1$, then
$$
\begin{array}{l}
2 \cdot 3^{y}-5^{x} \cdot 7^{x}=1 \\
\Rightarrow-5^{x} \cdot 7^{x} \equiv 1(\bmod 3) \\
\Rightarrow(-1)^{x} \equiv-1(\bmod 3) .
\end{array}
$$

Thus, $z$ is odd, so
$$
2 \cdot 3^{y}=1(\bmod 5) \text {. }
$$

This implies $y \equiv 1(\bmod 4)$.
If $w \neq 0$, then $2 \cdot 3^{y} \equiv 1(\bmod 7)$.
Therefore, $y \equiv 4(\bmod 6)$, which contradicts $y \equiv 1(\bmod 4)$.
Thus, $w=0$.
Hence, $2 \cdot 3^{y}-5^{z}=1$.
For $y=1$, $z=1$.
For $y \geqslant 2$, we have $5^{x} \equiv-1(\bmod 9)$, which implies $z \equiv 3(\bmod 6)$.
Therefore, $\left(5^{3}+1\right) \mid\left(5^{x}+1\right)$.
Thus, $71\left(5^{x}+1\right)$, which contradicts $5^{x}+1=2 \cdot 3^{y}$.
Thus, the solution in this case is
$$
(x, y, z, w)=(1,1,1,0) \text {. }
$$

Case 3 If $y>0, x \geqslant 2$, then
$$
5^{2} \cdot 7^{m} \equiv-1(\bmod 4), 5^{r} \cdot 7^{m} \equiv-1(\bmod 3),
$$

which implies $(-1)^{\infty} \equiv-1(\bmod 4),(-1)^{2} \equiv-1(\bmod 3)$.
Thus, $z$ and $w$ are both odd. Therefore,
$$
2^{x} \cdot 3^{y}=5^{x} \cdot 7^{x}+1 \equiv 35+1 \equiv 4(\bmod 8) \text {. }
$$

Thus, $x=2$.
The original equation becomes
$4 \cdot 3^{y}-5^{z} \cdot 7^{w}=1$ (where $z$ and $w$ are both odd).
This implies $4 \cdot 3^{y} \equiv 1(\bmod 5), 4 \cdot 3^{y} \equiv 1(\bmod 7)$.
From the above two equations, we get $y \equiv 2(\bmod 12)$.
Let $y=12 m+2, m \geqslant 0$. Thus,
$$
5^{2} \cdot 7^{m}=4 \cdot 3^{y}-1=\left(2 \cdot 3^{6 m+1}-1\right)\left(2 \cdot 3^{6 m+1}+1\right) \text {. }
$$

Since $2 \cdot 3^{6 m+1}+1 \equiv 6 \cdot 2^{3 m}+1 \equiv 6+1 \equiv 0(\bmod 7)$, and $\left(2 \cdot 3^{6 m+1}-1,2 \cdot 3^{6 m+1}+1\right)=1$, we have
$$
51\left(2 \cdot 3^{6 m+1}-1\right) \text {. }
$$

Thus, $2 \cdot 3^{6 m+1}-1=5^{x}$,
$2 \cdot 3^{6 m+1}+1=7^{\infty}$.
If $m \geqslant 1$, then from (1) we get $5^{x} \equiv-1(\bmod 9)$.
From Case 2, we know this is impossible.
If $m=0$, then $y=2, z=1, w=1$.
Thus, the solution in this case is
$$
(x, y, z, w)=(2,2,1,1) \text {. }
$$

In summary, the non-negative integer solutions $(x, y, z, w)$ are
$$
(1,0,0,0),(3,0,0,1),(1,1,1,0),(2,2,1,1) \text {. }
$$
(2005 China Mathematical Olympiad Committee)