Solve the Diophantine equation $ (3^{n}-1)(3^{m}-1)=x^{r} $

Question

Is there a solution other than solution

\begin{equation*} (3^{2}-1)(3^{2}-1)=2^{6}=4^{3} \end{equation*}

of the Diophantine equation \begin{equation*} (3^{n}-1)(3^{m}-1)=x^{r} \end{equation*} for positive integers $ n, m, x, r $ such as $ n, m, x \geq 2 $ and $ r \geq 3 $ ?

Looks interesting from here...where did this problem come from? — Mike, Jul 02 '22 at 23:07
the diophantine equation \begin{equation} (a^{n}-1)(b^{m}-1)=x^{r} \end{equation} and Catalan's conjecture — Djamel Djamel, Jul 02 '22 at 23:14
@DjamelDjamel Note that for any $n = m$, with $m \ge 2$, we have a solution of $x = 3^m - 1$ and $r = 2$, with this matching all of your conditions (as your solution shows for the case of $n = m = 2$). I assume you're looking for other solutions where $m \ne m$. — John Omielan, Jul 02 '22 at 23:27
@DjamelDjamel I suggest you make that clear in your question text. — John Omielan, Jul 02 '22 at 23:31

Will Jagy · Answer 1 · 2022-07-03T01:07:31.633

these ought to be relatively rare, by Zsigmondy. However, in that theorem, we are not told the exponent on the new prime. So, for example, $3^5 - 1 = 2 \cdot 11^2.$ The new prime, $11,$ is squared, and we find $$ (3^2 - 1)(3^5 - 1) = 44^2 $$

Usually, a new prime is has exponent $1.$

Made a list, my program was able to completely factor and show the new prime(s) until $n$ became a prime over 35... I could do those in gp-pari

Sat Jul  2 17:56:42 PDT 2022
1    2 =  2
2    4 = 2^2
3    13 =  13
4    5 =  5
5    121 = 11^2
6    7 =  7
7    1093 =  1093
8    41 =  41
9    757 =  757
10    61 =  61
11    88573 = 23  3851
12    73 =  73
13    797161 =  797161
14    547 =  547
15    4561 =  4561
16    3281 = 17  193
17    64570081 = 1871  34511
18    703 = 19  37
19    581130733 = 1597  363889
20    1181 =  1181
21    368089 =  368089
22    44287 = 67  661
23    47071589413 = 47  1001523179
24    6481 =  6481
25    3501192601 = 8951  391151
26    398581 =  398581
27    387440173 = 109 433  8209
28    478297 = 29  16493
29    34315188682441 = 59 28537  20381027
30    8401 = 31  271
31    308836698141973 = 683 102673  4404047
32    21523361 =  21523361
33    2413941289 =  2413941289
34    32285041 = 103 307  1021
35    189150889201 = 71  2664097031
36    530713 =  530713
37    225141952945498681 =  cdot mbox{BIG} =  13097927  17189128703  
38    290565367 = 2851  101917
39    15040635637 = 313 6553  7333
40    42521761 =  42521761
41    18236498188585393201 = 83  cdot mbox{BIG}  = 83 2526913 86950696619 
42    97567 = 43  2269
43    164128483697268538813 = 431  cdot mbox{BIG} = 431  380808546861411923
44    3138105961 = 5501  570461
45    271983020401 = 181 1621  927001

We have also, that $1006003^2 \mid 3^{1006002}-1$. So one could try your idea with this number as well. Unfortunately there are manymanymany digits in the game ... :-) (I've tried factoring this at the "alpertron factoring" site https://www.alpertron.com.ar/ECM.HTM, but well ...) — Gottfried Helms, Oct 15 '22 at 23:26

Djamel Djamel · Answer 2 · 2022-07-03T12:11:19.393

0

Let $ d= \gcd(3^{n}-1; 3^{m}-1) $, so $ d = 3^{\gcd(n;m)}-1 $

solve the diophantine equation \begin{equation} (3^{n}-1)(3^{m}-1)=x^{2} \end{equation} equivalent solving the system a of two Diophantine equations as follows: \begin{equation} 3^{n}-1=d \: \alpha^{2} \\ 3^{m}-1= d \: \beta^{2} \end{equation}

with $ \gcd(\alpha; \beta)= 1 $ and $ x= d \alpha \beta $.

edited Jul 03 '22 at 12:11

answered Jul 03 '22 at 11:38

Djamel Djamel

21

2

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Solve the Diophantine equation $ (3^{n}-1)(3^{m}-1)=x^{r} $

2 Answers2