Good day

The problem is as follow:

Find all solutions $(x, y)$, where $x, y \in \mathbb {Z^+}$ to the equation: $$1+3^x=2^y$$

Two solutions are $(0,1)$ and $(1,2)$ but how do you go about calculating other possible solutions or proving those are the only ones? Also, what would the general approach be to solving such an equation? Without a calculator, as you would need to do in an olympiad.

Thank you!


1 Answer 1


Hint. Examine the possible values modulo $8$ of $ 1 + 3^x$.

  • $\begingroup$ After doing this I found: 1+3^2k=2 (mod 8); 1+3^(2k+1)=4 (mod 8). 3^2k=1 (mod 8); 3^(2k+1)=3 (mod 8). 2^1=2 (mod 8); 2^2=4 (mod 8); 2^y=0 (mod 8) for y>2. When x=2k: 1+3^2k=2 (mod 8); we can then easily show y=1. Same for x=2k+1; y=2. It is not necessary to show any further, from 2^y=0 (mod 8). The only solution satisfying the problem's conditions thus is x=1, y=2. How does one select which modulo to use in such a problem? Thank you! $\endgroup$
    – user253171
    Commented Jul 16, 2015 at 22:38

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