I was working on this problem and came to a standstill. I'm not exactly sure how to go about this problem, to find if any integer pairs of $(x,y)$ satisfy this equation. Any guidance would be appreciated!

$$ y=\log_2(1+3^x) $$

  • 2
    $\begingroup$ $x=1$, $y=2$ is an obvious solution. $\endgroup$
    – Bernard
    May 28 '18 at 22:20
  • 1
    $\begingroup$ as is $x = 0$, $y = 0$ $\endgroup$ May 28 '18 at 22:23
  • $\begingroup$ oh I should have put that I meant for large values of x. My mistake $\endgroup$
    – wjmccann
    May 28 '18 at 22:27

Note that since $1 + 3^x > 1$, we have that $y$ is positive.

We need that $2^y = 3^x + 1$.

If $y \geq 3$, then $3^x + 1$ must be a multiple of $8$. But $3^x + 1$ is always either $2$ more or $4$ more than a multiple of $8$ (depending on whether $x$ is even or odd), so there are no solutions in this case.

So the only possible solutions occur when $0 < y < 3$.

  • $\begingroup$ Perfect thank you! $\endgroup$
    – wjmccann
    May 28 '18 at 22:28

The Catalan conjecture, that the only perfect powers differing by $1$ are $2^3=8$ and $3^2=9$, was resolved by Preda Mihăilescu in 2002. The question you're asking is equivalent to finding solutions to $2^y-3^x=1$, which is a special case of Catalan. There are no non-trivial solutions.


You don't actually need the full strength of Catalan's conjecture to show this particular result. When the bases are $2$ and $3$, the proof is simply a bit of finite group theory, accessible to anyone who knows about cyclic groups and induction.


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