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}),


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.)


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One response to “Catalan-type diophantine equations

  1. Pingback: Proof of a special case of Pillai’s conjecture | Samin Riasat's Blog

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