Number theory powers of numbers proof

• Aug 31st 2010, 05:03 PM
mel240
Number theory powers of numbers proof
Find all positive integers k and l such that 3^{k} - 2^{l} = 1

This is the last part of a question and I have previously proved that 2^{l} + 1 is divisible by 27 iff it is also divisible by 19 and did this by examining the sequence of powers of 2 reduced modulo 27 and the sequence of powers of 2 reduced modulo 19.
I'm assuming this somehow relates to this proof but not sure how to start?
• Aug 31st 2010, 06:25 PM
Khonics89
Well if you refer to the tables you drew out from parts i,ii...

you will notice some connection between them.. write 27 as 3^3 so

Also, look at how 3^k = 1+ 2^l

we can see when k=1, l=1 it gives us a solution..

Such a small world!
• Aug 31st 2010, 08:55 PM
simplependulum
$\displaystyle k=l=1$ is a solution and suppose $\displaystyle l \geq 2$ so $\displaystyle 2^l \equiv 0 \bmod{4}$ . Take modulo $\displaystyle 4$ , we have :

$\displaystyle (-1)^k \equiv 1 \bmod{4}$ which forces $\displaystyle k$ to be even .

Let $\displaystyle k = 2x$ and consider the factorization :

$\displaystyle 3^{2x} - 1 = 2^l = (3^x - 1)(3^x + 1 )$ . We have to set both of $\displaystyle 3^x - 1$ and $\displaystyle 3^x + 1$ to be the power of two because of the other side . Call them $\displaystyle 2^s$ and $\displaystyle 2^t$ respectively , so we have $\displaystyle 2^t - 2^s = 2$ , the only solution to this easy equation is $\displaystyle t=2 ~,~ s=1$ which gives $\displaystyle k=2 ~,~ l = 3$ the only solution other than $\displaystyle (k,l) = (1,1)$ .