Congruent equality - how to prove?

[AntonVrba]

If p is a prime or psuedo-prime base b then the following equality holds for all values \mbox{p , b}, but how to prove it?

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

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

Your help would be appreciated - thanks and regards

 

[ramsey2879]

If p is a prime or psuedo-prime base b then the following equality holds for all values \mbox{p , b}, but how to prove it?

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

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

Your help would be appreciated - thanks and regards

This reminds me of a math Joke

The less you know, the more you make.
Proof:

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 theorem?

 

[AntonVrba]

This reminds me of a math Joke

The less you know, the more you make.
...
Thus, as Knowledge approaches zero, Money approaches infinity,
regardless of the amount of Work done.


Do you know Fermat's Little theorem?

Love your joke - by your theory I should be rich - alas - far from it .

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