<s>[INST] 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 $y\ge 3$.
 [/INST] 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 $(x, y, z, w) = (2, 2, 1, 1)$.
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