# Catalan-type diophantine equations

Let’s say we want to find all integers $x,y$ such that

$2^x-5\cdot 3^y=1$.

Suppose that $x,y$ are sufficiently large. Having spotted the solution $x=4,y=1$, re-write the equation as

$2^x-5\cdot 3^y=2^4-5\cdot 3\Leftrightarrow 16(2^{x-4}-1)=3\cdot 5(3^{y-1}-1)$.

So $16\mid 3^{y-1}-1$. Checking the powers of 3 modulo 16 one deduces that $4\mid y-1$. Writing $y-1=4k$ therefore gives

$16(2^{x-4}-1)=3\cdot 5\cdot 80(1+81+\cdots+81^{k-1})$,

i.e.,

$2^{x-4}-1=3\cdot 5^2(1+81+\cdots+81^{k-1})$.

Hence $5^2=25\mid 2^{x-4}-1$. As before, one deduces that $20\mid x-4$. (A better way: since $\hbox{ord}_5(2)=4$, we have $\hbox{ord}_{25}(2)=5\cdot 4=20$ by this.) Writing $x-4=20m$ gives

$11\cdot 31\cdot 41(1+2^{20}+\cdots+2^{20(m-1)})=1+81+\cdots+81^{k-1}$.

Using $81\equiv -1\pmod{41}$, it follows that $k$ must be even. Then $1+81=82$ divides the left-hand side, which is impossible since the left-hand side is always odd. Thus we conclude that $x=4,y=1$ is the only solution.

This kind of approach usually works for many similar equations. The basic idea is the following:

1. Guess an initial solution.
2. Use the initial solution to group terms and factor them.
3. Use congruence relations to restrict the variables.
4. Eventually one side will likely have a prime factor that does not divide the other side.

This motivates the following:

Conjecture. Given integers $A, B, C, D, E>0$ with $BD>1$, the equation

$A\cdot B^x-C\cdot D^y=E$

has only finitely many solutions in integers.

This turns out to be a special case of a more general conjecture known as Pillai’s conjecture.

(All this arose when I was trying to make $A,C$ as small as possible with respect to $x,y$, which range through the positive integers satisfying

$A\cdot 2^x-C\cdot 3^y=1$.)