Post Number:#4 by ahmedittihad » Mon Feb 20, 2017 5:51 pm
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 $3^2(3^{m-2} - 1) = 7(7^{n-1}-1)$.
We have $3^2 \mid 7^{n-1}-1$; the multiplicative order of $7$ modulo $3^2$ is $3$, so we need $3\mid n-1$ and then $19\mid 7^3-1 \mid 7^{n-1}-1$.
We have $19 \mid 3^{m-2}-1$; the multiplicative order of $3$ modulo $19$ is $18$, so we need $3\mid n-1$ and then $37\mid 3^{18}-1 \mid 3^{m-2}-1$.
We have $37 \mid 7^{n-1}-1$; the multiplicative order of $7$ modulo $37$ is $9$, so we need $9\mid n-1$ and then $3^3\mid 7^9-1 \mid 7^{n-1}-1$.
But $3^3 \mid 3^2(3^{m-2} - 1)$ implies $3 \mid 3^{m-2} - 1$, impossible. Thus there are no other solutions.
Frankly, my dear, I don't give a damn.