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lecture12b

# lecture12b - CS 70 Spring 2005 Discrete Mathematics for CS...

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CS 70 Discrete Mathematics for CS Spring 2005 Clancy/Wagner Notes 12 RSA and the Chinese remainder theorem The Chinese remainder theorem Suppose we have a system of simultaneous equations, like maybe this one: x 2 ( mod 5 ) x 5 ( mod 7 ) What can we say about x ? Well, notice that one solution is x = 12; x = 12 satisfies both equations. This is not the only solution: for instance, x = 12 + 35 also works, as does x = 12 + 70, x = 12 + 105, and so on. Evidently adding any multiple of 35 to any solution gives another valid solution, so we might as well summarize this state of affairs by saying that x 12 ( mod 35 ) is one solution of the above system of equations. What about other solutions? For this example, there are no other solutions; every solution is of the form x 12 ( mod 35 ) . Why not? Well, suppose x and x 0 are two valid solutions. From the first equation, we know that x 2 ( mod 5 ) and x 0 2 ( mod 5 ) , so we must have x x 0 ( mod 5 ) . Similarly x x 0 ( mod 7 ) . But the former means that 5 is a divisor of x - x 0 , and the latter means that 7 is a divisor of x - x 0 , so x - x 0 must be a multiple of 35 (here we have used that gcd ( 5 , 7 ) = 1), which in turn means that x x 0 ( mod 35 ) . In other words, all solutions are the same modulo 35: or, equivalently, if all we care about is x mod 35, the solution is unique. You can check that the same would be true if we replaced the numbers 5 , 7 , 2 , 5 above by any others. The only thing we used is that gcd ( 5 , 7 ) = 1. Here is the generalization:

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lecture12b - CS 70 Spring 2005 Discrete Mathematics for CS...

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