# Are there integer remedies to $9^x - 8^y = 1$?

The only remedy is $x=y=1$ This is a grandfather clause of Catalan's conjecture which was confirmed by Mihailescu in 2002: the only remedy in all-natural varieties of $x^a - y^b = 1$ with $a,b\gt 1$ is $x=3$, $a=2$, $y=2$, and also $b=3$.

Yet as Moron mentions, it can be reasoned far more elementarily.

Except $x=1$ and also $y=1$ there aren't any kind of.

We have that

$$3^{2x} - 1 = 8^y$$

i.e

$$ (3^x + 1)(3^x - 1) = 8^y$$

Thus we have to have that

$$3^x + 1 = 2^m, 3^x - 1 = 2^n$$

Thus $$2^m - 2^n = 2$$

i.e. $$ 2^n(2^{m-n} - 1) = 2$$

Thus $n=1$ and also $m=2$.

Equation $\rm\ 3^{2x}-2^{3y}=1\ $ is an instance of numerous grandfather clauses of Catalan's Conjecture.

First, $\ $ making the field of expertise $\rm\ \ \: z,\:p^n = 3^x,2^{3y}\ $ listed below returns $\rm\ x = 1 = y\ $ as wanted.

**LEMMA **$\ \ $ $\rm z^2 - p^n = 1\ \ \Rightarrow\ \ z,\:p^n = \:3\ ,\:2^3\ $ or $\ 2,\:3\ $ for $\rm\ \ z,\:p\:,n\in \mathbb N,\ \ p\: $ prime

**Proof ** $\rm\ \ \ (z+1)\:(z-1)\: =\: p^n\ \ \Rightarrow\ \ z+1 = p^{\:j},\ \ z-1 = p^k\ $ for some $\rm\ j,\:k\in \mathbb N$

$\rm\quad \:\Rightarrow\ \ \ \ 2\ =\ p^{\:j} - p^k\ =\ p^k\: (p^{\:j-k}-1) \ \Rightarrow\ p^k=2\ $ or $\rm\ p^k = 1 \ \Rightarrow\ \ldots$

Second, it is merely the grandfather clause $\rm\: X = 3^x,\ Y = 2^y\: $ of $\rm\ X^2 - Y^3 = 1\:,\: $ addressed by Euler in 1738. Nowadays one can offer this remedy fairly conveniently making use of primary buildings of $\rm\ \mathbb Z[\sqrt\[3\]{2}]\:$, e.g see p. 44 of Metsankyla: Catalan's Conjecture: another old diophantine problem solved. See additionally this MO thread and also this MO thread and also Schoof: Catalan's Conjecture. Keep in mind additionally that Catalan formulas are a grandfather clause of the concept of generalised Fermat (FLT) formulas, as an example see Darmon's exposition.

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