- #1

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[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

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- Thread starter AntonVrba
- Start date

- #1

- 92

- 0

[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

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This reminds me of a math JokeAntonVrba said:

[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

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 Theorm?

- #3

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ramsey2879 said: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 Theorm?

Love your joke :rofl: - 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

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