Vtyjkyuk

Prove that $$2^x cdot 3^y - 5^z cdot 7^w = 1$$ has no solutions in $mathbbZ^+$, if $yge 3$.

Consider any rational number $2^x 3^y 5^-z 7^-w$, where $x$, $y$, $z$, $w in mathbbZ^+$ . 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:

$frac76, frac87, frac1514, frac2120, frac2827, frac3635, frac4948, frac5049, frac6463, frac126125, frac225224, frac24012400, frac43754374$

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 $frac3635$, or $(x, y, z, w) = (2, 2, 1, 1)$.

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
$$1=2^xcdot3^y-5^zcdot7^wequiv-(-1)^zpmod3.$$

Observation 2: $x=2$ and $yequiv2pmod12$.

Reducing mod $8$ shows that if $xgeq3$ then
$$1=2^xcdot3^y-5^zcdot7^wequiv-5^zcdot7^wpmod8,$$
which implies that $z$ is even, a contradiction, hence $xleq 2$. Reducing mod $5$ and $7$ shows that
$$1equiv2^xcdot3^yequiv2^x-ypmod5
qquadtext and qquad
1equiv2^xcdot3^yequiv3^2x+ypmod7,$$
which tells us that $x-yequiv0pmod4$ and $2x+yequiv0pmod6$. It follows that $y$ is even and hence also $x$ is even. Because $xleq2$ we find that $x=2$ and hence that $yequiv2pmod12$.

Observation 3: $w=1$.

Reducing mod $8$ shows that
$$1equiv4cdot3^y-5^zcdot7^wequiv4-5cdot7^wpmod8,$$
because $y$ is even and $z$ is odd, and so $w$ is also odd. Reducing mod $49$ shows that if $wgeq2$ then
$$1=4cdot3^y-5^zcdot7^wequiv4cdot3^ypmod49,$$
which implies that $yequiv32pmod42$. Then reducing mod $43$ shows that
$$1=4cdot3^y-5^zcdot7^wequiv9-5^zcdot7^wpmod43,$$
and hence $5^zcdot7^wequiv8pmod43$. 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 $ygeq3$ then
$$1=4cdot3^y-5^zcdot7equiv-5^zcdot7pmod27,$$
which implies that $zequiv13pmod18$. Then $5^zequiv17pmod19$ so reducing mod $19$ then shows that
$$1=4cdot3^y-5^zcdot7equiv4cdot3^y-17cdot7equiv4cdot3^y-5pmod19,$$
and so $4cdot3^yequiv6pmod19$. But $4cdot3^yequiv5,16,17pmod19$ because $yequiv2pmod6$, a contradiction, hence $yleq2$. Because $y$ is even we find that $y=2$.

Conclusion: The only solution to
$$2^xcdot3^y-5^zcdot7^w=1,$$
in the positive integers is $(x,y,z,w)=(2,2,1,1)$.