Solve the following equation in natural numbers:
$3^m7^n=2$.
Source: AOPS
Exponential Diophantine Eq
 Raiyan Jamil
 Posts: 138
 Joined: Fri Mar 29, 2013 3:49 pm
Re: Exponential Diophantine Eq
I think n=2 and m=1 . Am I correct?
A smile is the best way to get through a tough situation, even if it's a fake smile.

 Posts: 107
 Joined: Sun Dec 12, 2010 10:46 am
Re: Exponential Diophantine Eq
Proof please.
 ahmedittihad
 Posts: 174
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Exponential Diophantine Eq
There are no smaller solutions than $(m,n)=(2,1)$ , when $3^2  7^1 = 2$, so let's look for larger ones.
Write the equation as $3^2(3^{m2}  1) = 7(7^{n1}1)$.
We have $3^2 \mid 7^{n1}1$; the multiplicative order of $7$ modulo $3^2$ is $3$, so we need $3\mid n1$ and then $19\mid 7^31 \mid 7^{n1}1$.
We have $19 \mid 3^{m2}1$; the multiplicative order of $3$ modulo $19$ is $18$, so we need $3\mid n1$ and then $37\mid 3^{18}1 \mid 3^{m2}1$.
We have $37 \mid 7^{n1}1$; the multiplicative order of $7$ modulo $37$ is $9$, so we need $9\mid n1$ and then $3^3\mid 7^91 \mid 7^{n1}1$.
But $3^3 \mid 3^2(3^{m2}  1)$ implies $3 \mid 3^{m2}  1$, impossible. Thus there are no other solutions.
Write the equation as $3^2(3^{m2}  1) = 7(7^{n1}1)$.
We have $3^2 \mid 7^{n1}1$; the multiplicative order of $7$ modulo $3^2$ is $3$, so we need $3\mid n1$ and then $19\mid 7^31 \mid 7^{n1}1$.
We have $19 \mid 3^{m2}1$; the multiplicative order of $3$ modulo $19$ is $18$, so we need $3\mid n1$ and then $37\mid 3^{18}1 \mid 3^{m2}1$.
We have $37 \mid 7^{n1}1$; the multiplicative order of $7$ modulo $37$ is $9$, so we need $9\mid n1$ and then $3^3\mid 7^91 \mid 7^{n1}1$.
But $3^3 \mid 3^2(3^{m2}  1)$ implies $3 \mid 3^{m2}  1$, impossible. Thus there are no other solutions.
Frankly, my dear, I don't give a damn.