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    (Original post by JamesF)
    If you dont know about modulo arithmetic then you could do it like this instead, the same, but not as neat.
    4 = 3+1
    4^n = 3^n + ....(nCr)3^r.......+ 1, so each term apart from the last is a multiple of 3. Take the 1 away and your left with a multiple of 3.
    Why not generalise it then:

    (n^m - 1 ) mod (n-1) = 0
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    (Original post by figgetyfig)
    lol, I feel a little stupid looking at all this, because I immediately thought of induction, and not any of the other methods. Anyway, I know this has been done, but just to prove that I can...

    assume true for n=k
    4^k -1 = 3M (where M is an integer)
    therefore 4^k = 3M+1

    for n=k+1,
    4^(k+1) = 4x4^k
    therefore 4^(k+1) -1 = 4(3M+1) -1
    = 12M+3
    = 3(4M + 1)
    which is divisible by 3.

    If true for n=k, then true for n=k+1
    when n=1, 4^n -1 = 1-1 = 0, which is divisible by 3 (even though its not a multiple)
    when n=2, 4^n -1 = 16-1 = 15, which is divisible by 3.
    therefore true for all positive integer values of n.
    Yeh same here.............induction came straight to my head too, i didn't dare think of the way these other peeps did it!!!
 
 
 
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Updated: July 10, 2004
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