# Thread: Number theory powers of numbers proof

1. ## 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?

2. 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!

3. $\displaystyle k=l=1$ 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 $\displaystyle (k,l) = (1,1)$ .