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A few days ago I asked for a solution for $3^a+1=2^b$ where $a\in \Bbb N$ and $b\in \Bbb N$ (Solutions for diophantine equation $3^a+1=2^b$), and I got two good answers for this question.

But there I also asked for the more general case $p_1^a+n=p_2^b$ where $p_1$ and $p_2$ are prime and $a,b,n\in \Bbb N$. But this part of my question was ignored, so here I explicitly asks for ways to find solutions for this general case.


I still am interested in any $p_1$ and $p_2$, but im very focused on $p_1=3$ and $p_2=2$ because this is part of another problem that I try to solve.

I explicitly am interested in solutions for this equations:

  • $3^a+5=2^b$
  • $3^a+7=2^b$
  • $3^a+13=2^b$
  • $3^a+23=2^b$
  • ...

where $a\in \Bbb N$ and $b\in \Bbb N$ and where $b>2$ ($2^b>4$)

Supposed that $b>2$, you can show, that in

$2^b-3^a \equiv n \pmod{24}$

$n$ only can be $5,7,13$ or $23$

Proof:
$2^b \pmod{8}$ is always $0$ when $b>2$; $3^a \pmod{8}$ is always $1$ or $3$; so $2^b-3^a \pmod{8}$ only can be $5$ or $7$. $3^a \pmod{3}$ is always $0$ when $a>0$, but $2^b \pmod{3}$ is never $0$, it's either $1$ or $2$, so also $2^b-3^a \pmod{3}$ only can be $1$ or $2$. Together this leads to the fact, that $2^b-3^a \pmod{24}$ only can be $5,7,13$ or $23$ when $b>2$.
qed

Knowing this, I was searching for differences $n$ of powers of $2$ minus powers of $3$ having $n<100$ and I found this:

  • $n=5$:
    $3^1+5=2^3 \rightarrow 3+5=8$
    $3^3+5=2^5 \rightarrow 27+5=32$

  • $n=7$:
    $3^2+7=2^4 \rightarrow 9+7=16$

  • $n=13$:
    $3^1+13=2^4 \rightarrow 3+13=16$
    $3^5+13=2^8 \rightarrow 243+13=265$

  • $n=23$:
    $3^2+23=2^5 \rightarrow 9+23=32$

  • $n=29 = 5+24$:
    $3^1+29=2^5 \rightarrow 3+29=32$

  • $n=31 = 7+24$:
    no solution found

  • $n=37 = 13+24$:
    $3^3+37=2^6 \rightarrow 27+37=64$

  • $n=47 = 23+24$:
    $3^4+47=2^7 \rightarrow 81+47=128$

  • $n=53 = 5+2*24$:
    no solution found

  • $n=55 = 7+2*24$:
    $3^2+55=2^6 \rightarrow 9+55=64$

  • $n=61 = 13+2*24$:
    $3^1+61=2^6 \rightarrow 3+61=64$

  • $n=71,77,79,85,95$:
    no solutions found

By comparing any power of 2 up to a certain limit (which was $b \le 2^{10}$) with the biggest powers of 3 being smaller than that power of 2, I found out, that all other differences of powers of 2 minus powers of 3 are bigger than 100 for all $b \le 2^{10}$ which means $2^b \le 1.8 \times 10^{307}$.

So the solutions for $n<100$ listed above are the only existing solutions with $3^a$ and $2^b$ having less than 307 decimal digits. And this makes me believe, that there also are no solutions for $n=5,7,13,23,31,...,95$ when $3^a$ and $2^b$ are bigger than $10^{307}$.

And I also think, that there is no difference, that appears infinitely often. I even think, that the maximum number that a given difference can appear is 2.

But I have no idea how to prove this.
Can you help?

  • Are there any other solutions for low values of $n$? If yes: How can you find them? If no: How can you prove that?
  • How many values for $a$ (or $b$) can share the same difference $n$?
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    $\begingroup$ One can devise a systematic method of finding all the solutions, since explicit lower bounds on $|2^a-3^b|$ can be given. See here $\endgroup$
    – Wojowu
    Commented Dec 24, 2016 at 14:51
  • $\begingroup$ @Wojowu: Thanks a lot! This article is very interesting. $\endgroup$ Commented Dec 24, 2016 at 15:22
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    $\begingroup$ I've found an article of Michael Bennett on the work of S. Pillai ( "Pillai's conjecture revisited" ). There are theorems which state that for $p_1=3,p_2=2$ or vice versa there is always at most only one solution for one specific difference (formulated as $|(n+1)^a - n^b|=d$ except for a very small list having 6 entries if $n=2$. So one can stop large quests for possible repeated solutions... (Also I've tried myself to solve the general case for $(p_1,p_2) = (3,2)$ with elementary means but am not yet sure whether my attempt allows also a general proof) $\endgroup$ Commented Dec 24, 2016 at 20:46
  • $\begingroup$ @GottfriedHelms: I read your profile. My native language is German (but I'm not German like you, I am from Austria). Maths is a hobby for me (as it is for you) and I came to this power of 2/3-thing via the Collatz Conjecture. So we seem to have something in common. But I have a problem: When I read English books about mathematics, I don't understand some basic terms just because they are named different in German. So please can you help me? Where can I find a book about number theory, that uses German and English terms? $\endgroup$ Commented Dec 24, 2016 at 22:08
  • $\begingroup$ Hubert, hi! - ganz mein Problem. Meine Tochter, eine 1+-englische, findet mein mathematisches Englisch ziemlich kauderwelschig, und ich habe in vieler HInsicht ähnliche Schwierigkeiten, komplexere englische Texte zu verstehen, geschweige selbst zu schreiben... ;-) Die Diskussionen in meiner eigentlichen deutschen "Heimat"-Newsgroup haben vor Jahren repide nachgelassen so daß ich mich wohl-oder-übel zu meinem half-baked Englisch gezwungen sehe (aber man hat ja dadurch auch ein größeres Publikum). Kann leider im Moment keine spezielle Buchempfehlung machen - aber mal sehen... $\endgroup$ Commented Dec 24, 2016 at 23:55

1 Answer 1

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CW answer, too long for comment. A student introduced an elementary method which is pretty good when the numbers do not get too large.

Exponential Diophantine equation $7^y + 2 = 3^x$

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.

Finding solutions to the diophantine equation $7^a=3^b+100$

I should add that @Gottfried Helms

https://math.stackexchange.com/users/1714/gottfried-helms

came up with a variation that has successfully handled some problems with bigger numbers; from the point of view of my answers, "big" means the size of the pair of primes that are used.

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