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A search for perfect squares

Understand the problem

Determine all pairs $(m,n)$ of positive integers for which $2^{m} + 3^{n}$ is a perfect square.

Source of the problem

Indian National Mathematical Olympiad 1992

Topic
Number Theory
Difficulty Level
Easy
Suggested Book
An Excursion in Mathematics

Start with hints

Do you really need a hint? Try it first!

First consider n=0. In this case the equation can be rewritten as 2^m=(x+1)(x-1). Hence, solving this case is equivalent to looking for powers of 2 that differ by 2.
For n\ge 1, consider residues modulo 3. Note that a perfect square cannot have remainder 2 upon division by 3.
Note that 2^m+3^n\equiv (-1)^m\;(\text{mod}\;3). From hint 3, it follows that m has to be even. Writing m=2p we get 3^n=(x+2^p)(x-2^p).
From hint 3, we may write x+2^p=3^i and x-2^p=3^j for some i and j. This gives 3^i-3^j=2^{p+1}. As the RHS is not divisible by 3, j has to be 0. Thus, 2^{p+1}=3^i-1. If p=0 then n=1. Otherwise, the LHS is 0 modulo 4, i.e. 1\equiv 3^i\equiv (-1)^i\;(\text{mod}\;4). Hence i=2k for some k. Now we again write 2^{p+1}=(3^k+1)(3^k-1). Again, we have two powers of 2 differing by 2. It is easy to see that (in case you have not noticed already) only such powers are 2 and 4. Hence 3^k-1=2.

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