China Mathematical Olympiad

Since $5^z\cdot 7^w+1$ is even, we have $x\ge 1$.

Case 1: $y=0$. The equation to be solved becomes $$2^x-5^z\cdot 7^w=1.$$ If $z\ne 0$, then $2^x\equiv 1(\mathrm{mod}5)$. It follows that $4\mid x$. Thus $3\mid 2^x-1$, which contradicts to $2^x-5^z\cdot 7^w=1$. If $z=0$, then $$2^x-7^w=1.$$ When $x=1,2,3$, a direct computation shows that $(x,w)=(1,0),(3,1)$ are the solutions. When $x\ge 4$, $7^w\equiv -1(\mathrm{mod}16)$. By direct computation we know that this is impossible. Consequently, when $y=0$ all non-negative integer solutions of the equation are $$(x,y,z,w)=(1,0,0,0),(3,0,0,1).$$

Case 2: $y>0$ and $x=1$. Thus the equation to be solved becomes $$2\cdot 3^y-5^z\cdot 7^w=1.$$ Hence $-5^z\cdot 7^w\equiv 1(\mathrm{mod}3)$, i.e., $(-1{)}^z\equiv -1(\mathrm{mod}3)$. It follows that $z$ is odd. Thus $$2\cdot 3^y\equiv 1(\mathrm{mod}5).$$ $$y\equiv 1(\mathrm{mod}4).$$ When $w\ne 0$, we have $2\cdot 3^y\equiv 1(\mathrm{mod}7)$. Thus $y\equiv 4(\mathrm{mod}6)$, which contradicts to the fact $y\equiv 1(\mathrm{mod}4)$. Hence $w=0$ and $$2\cdot 3^y-5^z=1.$$ When $y=1$, we have $z=1$. If $y\ge 2$, then $5^z\equiv -1(\mathrm{mod}9)$, which implies $z\equiv 3(\mathrm{mod}6)$. Thus $5^3+1(\mathrm{mod}{5}^z+1)$, so $7(\mathrm{mod}{5}^z+1)$, which contradicts to $5^z+1=2\cdot 3^y$. Hence in this case we have only one solution $$(x,y,z,w)=(1,1,1,0).$$

Case 3: $y>0$ and $x\ge 2$. Thus $$5^z\cdot 7^w\equiv -1(\mathrm{mod}4),\text{ and }{5}^z\cdot 7^w\equiv -1(\mathrm{mod}3).$$ That is, $$(-1{)}^w\equiv -1(\mathrm{mod}4),\text{ and }(-1{)}^z\equiv -1(\mathrm{mod}3).$$ Thus $z$ and $w$ are odd. It follows that $$2^x\cdot 3^y=5^z\cdot 7^w+1\equiv 35+1\equiv 4(\mathrm{mod}8).$$ Hence, $x=2$, and $$4\cdot 3^y-5^z\cdot 7^w=1\text{ (where z and w are odd).}$$ Thus, $$4\cdot 3^y\equiv 1(\mathrm{mod}5),\text{ and }4\cdot 3^y\equiv 1(\mathrm{mod}7).$$ From the above two congruencies we have $y\equiv 2(\mathrm{mod}12)$. Set $y=12m+2$, $m\ge 0$, then $$5^z\cdot 7^w=4\cdot 3^y-1=(2\cdot 3^{6m+1}-1)(2\cdot 3^{6m+1}+1).$$ Since $$2\cdot 3^{6m+1}+1\equiv 6\cdot 2^{3m}+1\equiv 6+1\equiv 0(\mathrm{mod}7),$$ and $$(2\cdot 3^{6m+1}-1,2\cdot 3^{6m+1}+1)=1,$$ so $5\mid 2\cdot 3^{6m+1}-1$. Thus $$2\cdot 3^{6m+1}-1=5^z,$$ $$2\cdot 3^{6m+1}+1=7^w.$$ If $m\ge 1$, by Equation (2) we have $5^z\equiv -1(\mathrm{mod}9)$, and from Case 2 we know that this is impossible. If $m=0$, then $y=2$, $z=1$ and $w=1$. Thus in this case, we have only one solution $$(x,y,z,w)=(2,2,1,1).$$

Consequently, all non-negative integer solutions are $$\begin{gathered} (x,y,z,w)=(1,0,0,0),(3,0,0,1), \\ (1,1,1,0),(2,2,1,1). \end{gathered}$$