Solutions for diophantine equation $3^a+1=2^b$

Question

I am looking for solutions for the diophantine equation

$3^a+1=2^b$

where $a\in \Bbb N$ and $b\in \Bbb N$.

Is there a power of $3$ that gives a power of $2$ when you add $1$?

Two solutions are easy to find:

$3^0+1=2^1 \rightarrow 1+1=2$
$3^1+1=2^2 \rightarrow 3+1=4$

But I'm looking for other solutions (solutions where $a>1$).

I believe that there is no other solution, but how can you proof this conjecture?

More general:
How can you find solutions for

$p_1^a+n=p_2^b$

where $p_1$ and $p_2$ are prime and $a,b,n\in \Bbb N$?

Following question addresses general case $3^a-2^b = \pm 1$ : $|2^x-3^y|=1$ has only three natural pairs as solutions — Sil, Dec 21 '16 at 11:13
More generally: Can we find positive integers $a$ and $k \geq 2$ with $2^n - 1 = a^k$? — Bart Michels, Dec 21 '16 at 11:13
@Sil there are \pm and \mp which output $\pm$ and $\mp$, respectively ;-) — Bart Michels, Dec 21 '16 at 11:14
By the way, thanks for contributing to the network of linked questions! — Bart Michels, Dec 21 '16 at 11:18
Btw, the equation $p_1^a+n=p_2^b$ can be considered as a particular case of Pillai's equation. You can check these links: http://math.stackexchange.com/questions/1642096/special-case-of-pillais-conjecture and http://mathoverflow.net/questions/29677/is-there-a-solution-for-the-equation-xm-yn-k-in-which-k-1 — Xam, Dec 21 '16 at 15:42

Roman83 · Accepted Answer · 2016-12-21T11:17:27.717

4

$$3^a=2^b-1$$ Let $b=2k, k \in \mathbb N, k \ge 1$ $$3^a=(2^k-1)(2^k+1)$$ Then

$$2^k-1=3^l$$ $$2^k+1=3^m$$ where $m+l=a, m>l$ Then $$3^m+3^l=2^{k+1}$$ $$3^l(3^{m-l}+1)=2^{k+1}$$ Then $l=0, k=1, m=1$

2) Let $b=2k+1$. Then $2^b≡2 (\mod 3)⟹a=0 $

edited Dec 21 '16 at 11:17

answered Dec 21 '16 at 11:10

Roman83

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Also, if $b=2k+1$, you have $2^b \equiv 2\ (\textrm{mod}\ 3) \implies a=0 $ – Baconaro Dec 21 '16 at 11:14
@Baconaro: Thank you! – Roman83 Dec 21 '16 at 11:17
I miss the general part of my question: How does your answer help me to find solutions for 3^a+5=2^b (27+5=32 is one solution for this)? – Hubert Schölnast Dec 21 '16 at 15:44

score 1 · Answer 2 · answered Dec 21 '16 at 15:10

Another solution: consider the powers of $3$$\pmod 8$. We have $3^1\equiv 3\pmod 8$ and $3^2\equiv 1\pmod 8$. So by induction it follows that $3^{2k+1}\equiv 3\pmod 8$ and $3^{2k}\equiv 1\pmod 8$, for $ k\in\Bbb {N} $. So $3^a+1$ would be congruent with $4$ or $2$$\pmod 8$. Therefore we deduce that if $3^a+1=2^b $, then $ b\le 2$.

If $ b=2$, then $ a=1$.

If $ b=1$, then $ a=0$. Hence all the solutions are the pairs $(a, b)=(0,1), (1,2) $.

Solutions for diophantine equation $3^a+1=2^b$

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