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!
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?
is a solution and suppose so . Take modulo , we have :
which forces to be even .
Let and consider the factorization :
. We have to set both of and to be the power of two because of the other side . Call them and respectively , so we have , the only solution to this easy equation is which gives the only solution other than .