I was thinking about different types of equations in two variables. My motivation for this is purely intrinsic. I just wanted to know what we can say about analytic methods of solving different equations. On the course of this, I came across the following:

$5^x - 3^y = 2$

My guess is that the only solution in the positive integers is $x=y=1$. Also Wolfram Alpha is telling me that. Is there a way of approaching this problem.

Of course we can write $y = \frac{\log(5^x - 2)}{\log(3)}$, but I am not sure that this helps.

Thank you very much for your help!

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    $\begingroup$ For integer solutions, I believe this is in general a very hard problem. However, for general solutions involving equations with exponents, maybe it will be helpful to check out the Lambert W Function $\endgroup$ – Michael Morrow Jun 8 at 7:58
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    $\begingroup$ I am not sure whether the negative result about the solvability of Hilbert's problem already occurs in two variables, but there is no easy way to solve them in general. $\endgroup$ – Peter Jun 8 at 8:03
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    $\begingroup$ I hope this answers your question en.wikipedia.org/wiki/Catalan%27s_conjecture $\endgroup$ – Maximal_inequality Jun 8 at 8:06
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    $\begingroup$ As far as I am aware , Catalan’s conjecture was proved in 2002 by Preda Mihăilescu. So the answer to your question is yes. $\endgroup$ – Maximal_inequality Jun 8 at 8:10
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    $\begingroup$ There is a conjecture that $x^a-y^b=n$ only has finite many solutions for every positive integer $n$ (Pillai's conjecture). But this has only be proven for $n=1$ (in this case $(9/8)$ is the only solution). However, the above equation is so special that someone will probably find a proof that $(1/1)$ is in fact the only solution. $\endgroup$ – Peter Jun 8 at 8:12

Firstly, we start with some modulus arguments. If we take the whole sequence modulo $3$, then we have that $5^x\equiv2$. In particular, $x\equiv1\pmod2$. Similarly, taking the sequence modulo $5$ tells us that $y\equiv1\pmod2$ as well.

The next step is to write $5^{x+1}=p^2$ and $3^{y-1}=q^2$. The equation then becomes, after multiplication by $5$, $$p^2-15q^2=10.$$ This is a generalised pell equation, for which the solutions are given here (Theorem 3.3). In particular, we solve the pell equation $$a^2-db^2=1$$ for $d=15$, which yields a primitive solution $(4,1)$, and then find all solutions to $$x^2-dy^2=n$$ for $n=10$ such that $|x|\le \sqrt{10(4+\sqrt{15})}\approx8.9$.

The only such solution, as you may have guessed, is $(5,1)$.

Finally, we compose the two solutions to find all solutions to the original equation, $$p^2-15q^2=10.$$ This is a simple recurrence, beginning with initial term $p_0=5,q_0=1$ with the conditions $$p_{n+1} = 4p_n + 15q_n,\\q_{n+1}=p_n+4q_n.$$

The trick to notice here is that this sequence can hardly ever have a power of $5$ as $p_n$ (which was, after all, the original question!) so we restrict ourselves to looking at powers of $5$. A straightforward induction tells us that $p_n$ satisfies $p_0=5,p_1=35,$ and $$p_{n+2}=8p_{n+1}-p_{n}$$ which is handy.

If you think about the condition that we don't want any other factors other than $5$ to appear, this sounds awfully reminiscent of Zsigmondy's Theorem -- indeed, the generalisation to Lucas sequences instantly reduces it to the finite case of $n\le30$.

>>> def is_power_of_5(n):
        while n%5==0:
        return n==1

>>> arr = [5,35]

>>> for i in range(40):

>>> for i in arr:
        if is_power_of_5(i):

Hence $(1,1)$ is indeed the only solution to the question.

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