Prove that $2^x \cdot 3^y - 5^z \cdot 7^w = 1$ has no solutions Prove that $$2^x \cdot 3^y - 5^z \cdot 7^w = 1$$ has no solutions in $\mathbb{Z}^+$, if $y\ge 3$.
 A: Consider any rational number $2^x 3^y 5^{-z} 7^{-w}$, where $x$, $y$, $z$, $w \in \mathbb{Z}^+$ .  Størmer's theorem guarantees that there are a finite number of such fractions where the numerator and denominator are consecutive integers.
Here are all of the solutions:
$\frac{7}{6}, \frac{8}{7}, \frac{15}{14}, \frac{21}{20}, \frac{28}{27}, \frac{36}{35}, \frac{49}{48}, \frac{50}{49}, \frac{64}{63}, \frac{126}{125}, \frac{225}{224}, \frac{2401}{2400}, \frac{4375}{4374}$
We only care about solutions where the numerator is $2^x 3^y$, and where the denominator is $5^z 7^w$ (for positive $x$, $y$, $z$, $w$).  The only fraction which satisfies this condition is $\frac{36}{35}$, or $(x, y, z, w) = (2, 2, 1, 1)$.
A: Here's a proof without Størmer's theorem, relying purely on (a lot of) modular arithmetic:
Observation 1: $z$ is odd.
Reducing mod $3$ shows that 
$$1=2^x\cdot3^y-5^z\cdot7^w\equiv-(-1)^z\pmod{3}.$$
Observation 2: $x=2$ and $y\equiv2\pmod{12}$.
Reducing mod $8$ shows that if $x\geq3$ then
$$1=2^x\cdot3^y-5^z\cdot7^w\equiv-5^z\cdot7^w\pmod{8},$$
which implies that $z$ is even, a contradiction, hence $x\leq 2$. Reducing mod $5$ and $7$ shows that
$$1\equiv2^x\cdot3^y\equiv2^{x-y}\pmod{5}
\qquad\text{ and }\qquad
1\equiv2^x\cdot3^y\equiv3^{2x+y}\pmod{7},$$
which tells us that $x-y\equiv0\pmod{4}$ and $2x+y\equiv0\pmod{6}$. It follows that $y$ is even and hence also $x$ is even. Because $x\leq2$ we find that $x=2$ and hence that $y\equiv2\pmod{12}$.
Observation 3: $w=1$.
Reducing mod $8$ shows that
$$1\equiv4\cdot3^y-5^z\cdot7^w\equiv4-5\cdot7^w\pmod{8},$$
because $y$ is even and $z$ is odd, and so $w$ is also odd. Reducing mod $49$ shows that if $w\geq2$ then
$$1=4\cdot3^y-5^z\cdot7^w\equiv4\cdot3^y\pmod{49},$$
which implies that $y\equiv32\pmod{42}$. Then reducing mod $43$ shows that
$$1=4\cdot3^y-5^z\cdot7^w\equiv9-5^z\cdot7^w\pmod{43},$$
and hence $5^z\cdot7^w\equiv8\pmod{43}$. But $5$, $7$ and $8$ are all quadratic nonresidues mod $43$, and $z$ and $w$ are odd, a contradiction, hence $w=1$.
Observation 4: $y=2$.
Reducing mod $27$ shows that if $y\geq3$ then
$$1=4\cdot3^y-5^z\cdot7\equiv-5^z\cdot7\pmod{27},$$
which implies that $z\equiv13\pmod{18}$. Then $5^z\equiv17\pmod{19}$ so reducing mod $19$ then shows that
$$1=4\cdot3^y-5^z\cdot7\equiv4\cdot3^y-17\cdot7\equiv4\cdot3^y-5\pmod{19},$$
and so $4\cdot3^y\equiv6\pmod{19}$. But $4\cdot3^y\equiv5,16,17\pmod{19}$ because $y\equiv2\pmod{6}$, a contradiction, hence $y\leq2$. Because $y$ is even we find that $y=2$.
Conclusion: The only solution to
$$2^x\cdot3^y-5^z\cdot7^w=1,$$
in the positive integers is $(x,y,z,w)=(2,2,1,1)$.
