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Congruent equality - how to prove?

  1. Sep 11, 2005 #1
    If [itex]p [/itex] is a prime or psuedo-prime base [itex]b [/itex] then the following equality holds for all values [itex]\mbox{p , b} [/itex], but how to prove it?

    [itex]b^{2p-4}\equiv\mbox{x (mod p)}\Longleftrightarrow b^{p-3}\equiv\mbox{x (mod p)}[/itex]

    we are talking about one and the same value for [itex]x [/itex] in the above two congruencies.

    Your help would be appreciated - thanks and regards
  2. jcsd
  3. Sep 11, 2005 #2
    This reminds me of a math Joke

    The less you know, the more you make.

    Postulate 1: Knowledge is Power.
    Postulate 2: Time is Money.
    As every engineer knows: Power = Work / Time
    And since Knowledge = Power and Time = Money
    It is therefore true that Knowledge = Work / Money .
    Solving for Money, we get:
    Money = Work / Knowledge
    Thus, as Knowledge approaches zero, Money approaches infinity,
    regardless of the amount of Work done.

    Another way to get this is that since equals divided by equals (not 0) are equal, then Money/Time = Power/Knowledge or
    Money = Power*Time/Knowledge = Work/Knowledge.

    Do you know Fermat's Little Theorm?
  4. Sep 12, 2005 #3
    Love your joke :rofl: - by your theory I should be rich - alas - far from it :frown: .

    Yes I know Fermats small theorem

    Mod(5^(13-3),13) = Mod(5^10,13) = 5
    Mod(5^(26-4),13) = Mod(5^22,13) = 5

    And Fermats Little theorom

    Mod(5^13,13) = 5 or Mod(5^12,13)=1

    Ok now I see - Mod(5^22,13)=Mod(5^10,13)*Mod(5^12,13) (a=b(mod c) implies k a=k b(mod c))

    Thanks for the joke
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