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 $n$?
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.