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This question already has an answer here:

Is there a way to find the solutions of $3^y=2^x-1$ where $(x,y)$ are integer coordinates (maybe within a certain interval)? Are there an infinite number of integer coordinate pairs for this equation? Is it possible to even determine this? I'm not quite sure the difficulty of this problem, but substituting $x$ or $y$ values into this equation to find integer pairs is time consuming and not very promising.

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marked as duplicate by Sil, GNUSupporter 8964民主女神 地下教會, José Carlos Santos, Strants, Mike Earnest Mar 22 at 17:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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This particular equation is a special case of Mihailescu's theorem: rewriting it as $2^x - 3^y = 1$, we know that either $x \le 1$ or $y \le 1$, because the only instance of higher powers having difference $1$ is $3^2 - 2^3$. This gives us the solutions $(x,y) = (2,1)$ and $(x,y) = (1,0)$.

In general, exponential Diophantine equations like this one can be solved by tedious application of the technique "take the equation modulo some arbitrary number like $37$ or something".

It's not quite that bad, though: we would just look for values of $m$ such that $2$ or $3$ have small multiplicative order mod $m$, i.e., $m$ divides $2^k-1$ or $3^k-1$ for some small $k$. This would either tell us that no solutions exist or give us some modular conditions on $x$ and $y$ which form part of a contradiction.

If some small solutions do exist, we would have to consider the equation modulo $2^k$ or $3^k$ to eliminate larger solutions. For example, in this problem, we have solutions for $x=1$ and $x=2$. If we had a solution in which $x \ge 3$, we would have $2^x \equiv 0 \pmod 8$. In that case, $$3^y \equiv 2^x - 1 \equiv 7 \pmod 8$$ but this is impossible: $3^y \bmod 8$ is $3$ when $y$ is odd, and $1$ when $y$ is even. So there can be no solution with $x \ge 3$.

(It's easy to check $x=0$, $x=1$, and $x=2$ to figure out that the above solutions are the only ones possible.)

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According to Catalan's conjecture (or Mihăilescu's theorem), there are no non-trivial solutions. The theorem states that the only solution in the natural numbers of $x^a − y^b = 1$ for $a, b > 1, x, y > 0$ is $x = 3, a = 2, y = 2, b = 3.$ link for the wiki page.

Despite of this powerful theorem, there is an easier way to solve this specific equation.

If $y=1$, then $x=2$. If $y >1$, then $6|x$ by taking modulo $9$. This will imply $2^x-1$ is divisible by $7$.

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  • $\begingroup$ The point here is not to quote some body else's answer. Try your own first.Even though this is an old question ( decades old ), it will be much more refreshing to come up with "another" answer than using existing answer. $\endgroup$ – DeepSea Apr 8 '17 at 22:59
  • $\begingroup$ @DeepSea: there are valid objections to answers that contain only a link, but your advice to userabc is unfair to a new user. You could more constructively have pointed to the answer by Misha Lavrov that gives the link to the key reference but also gives some extra useful information. Suggesting that people answering questions should always do so from their own work is absurd. $\endgroup$ – Rob Arthan Apr 8 '17 at 23:18
  • $\begingroup$ Thanks for the advice. Is it better for me to just leave a comment rather than an answer in this case? $\endgroup$ – userabc Apr 8 '17 at 23:24
  • $\begingroup$ The bad thing is not answering with a link, but not providing context: see "How do I write a good answer?". In this case, it would definitely be better to state the theorem and how it applies, then link to Wikipedia, but then that would be a good answer. (I decided to elaborate further because I felt that this question was not just "How do I solve this equation?" but "How do I solve equations like this one?", not because citing theorems is verboten.) $\endgroup$ – Misha Lavrov Apr 8 '17 at 23:49
  • $\begingroup$ I appreciate all the advice and will keep that in mind. Let me edit my answer to make it better. $\endgroup$ – userabc Apr 9 '17 at 0:12
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There is an entirely elementary way to show that $(x,y)=(1,0)$ and $(2,1)$ are the only integer solutions to $3^y=2^x-1$.

If $y\ge1$, $x$ must be even since $2^{2u+1}-1\equiv1$ mod $3$. Writing $x=2u$, we have

$$3^y=2^{2u}-1=(2^u-1)(2^u+1)$$

which implies $2^u-1$ and $2^u+1$ are each a power of $3$, say $2^u-1=3^r$ and $2^u+1=3^{r+s}$ with $s\ge1$ (since $2^u+1$ is clearly greater than $2^u-1$). This gives us

$$2=(2^u+1)-(2^u-1)=3^r(3^s-1)$$

the only solution to which is $r=0$ and $s=1$, leading to $u=1$, which is to say $(x,y)=(2,1)$.

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