17. (CMO-20 Problem) Find all non-negative integer solutions $(x, y, z, w)$ to the equation $2^{x} \cdot 3^{y}-5^{z} \cdot 7^{w}=1$.
17. From $5^{x} \cdot 7^{w}+1$ being even, we know $x \geqslant 1$.

Case I: If $y=0$, at this time $2^{x}-5^{x} \cdot 7^{w}=1$.
If $z \neq 0$, then $2^{x} \equiv 1(\bmod 5)$, which implies $4 \mid x$. Therefore, $3 \mid\left(2^{x}-1\right)$, which contradicts $2^{x}-5^{x} \cdot 7^{w}=1$.
If $z=0$, then $2^{x}-7^{w}=1$.
When $x=1,2,3$, direct calculation yields $(x, w)=(1,0),(3,1)$.
When $x \geqslant 4$, we have $7^{w} \equiv 1(\bmod 16)$, direct calculation shows this is impossible.
Therefore, when $y=0$, all non-negative integer solutions are $(1,0,0,0),(3,0,0,1)$.
Case II: If $y>0, x=1$, then $2 \cdot 3^{y}-5^{z} \cdot 7^{w}=1$.
Thus, $-5^{z} \cdot 7^{w} \equiv 1(\bmod 3)$, i.e., $(-1)^{z} \equiv-1(\bmod 3)$, hence $z$ is odd, so $2 \cdot 3^{y} \equiv 1(\bmod 5)$, which implies $y \equiv 1(\bmod 4)$.

When $w \neq 0$, we have $2 \cdot 3^{y} \equiv 1(\bmod 7)$, thus $y \equiv 4(\bmod 6)$. This contradicts $y \equiv 1(\bmod 4)$, so $w=0$. Therefore, $2 \cdot 3^{y}-5^{z}=1$.

When $y=1$, $z=1$; when $y \geqslant 2$, we have $5^{x} \equiv-1(\bmod 9)$, which implies $z \equiv 3(\bmod 3)$, thus $\left(5^{3}+1\right) \mid\left(5^{z}+1\right)$, hence $7 \mid\left(5^{z}+1\right)$, which contradicts $5^{z}+1=2 \cdot 3^{y}$. Therefore, the solution in this case is $(1,1,1,0)$.

Case III: If $y>0, x \geqslant 2$, by taking the original equation modulo 4 and modulo 3, we get $5^{z} \cdot 7^{w} \equiv-1(\bmod 4), 5^{z} \cdot 7^{w} \equiv-1(\bmod 3)$, i.e., $(-1)^{w} \equiv-1(\bmod 4),(-1)^{z} \equiv-1(\bmod 3)$, thus $z$ and $w$ are both odd, hence $2^{x} \cdot 3^{y}=5^{z} \cdot 7^{w}+1 \equiv 35+1 \equiv 4(\bmod 8)$, so $x=2$. The original equation becomes $4 \cdot 3^{y}-5^{z} \cdot 7^{w}=1$ (where $z$ and $w$ are both odd), from which we know $4 \cdot 3^{y} \equiv 1(\bmod 5), 4 \cdot 3^{y} \equiv 1(\bmod 7)$.
From the above two equations, we get $y \equiv z(\bmod 12)$. Let $y=12 m+2, m \geqslant 0$, then,
$5^{z} \cdot 7^{w}=4 \cdot 3^{y}-1=\left(2 \cdot 3^{6 m+1}-1\right)\left(2 \cdot 3^{6 m+1}+1\right)$.
Since $2 \cdot 3^{6 m+1}+1 \equiv 6 \cdot 2^{3 m}+1 \equiv 6+1 \equiv 0(\bmod 7)$,
and since $\left(2 \cdot 3^{6 m+1}-1,2 \cdot 3^{6 m+1}+1\right)=1$, so $5 \mid\left(2 \cdot 3^{6 m+1}-1\right)$, thus
$2 \cdot 3^{6 m+1}-1=5^{z}, 2 \cdot 3^{6 m+1}+1=7^{w}$.
If $m \geqslant 1$, then from the above equations, we get $5^{z} \equiv-1(\bmod 9)$, which has been proven impossible in Case II.
If $m=0$, then $y=2, z=1, w=1$. The solution in this case is $(2,2,1,1)$.
In summary, all non-negative integer solutions are $(1,0,0,0),(3,0,0,1),(1,1,1,0),(2,2,1,1)$.