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The Euler Phi-Function Is Multiplicative (3/3) Last time we showed that if p is prime, then  (p k ) = p k – p k-1. Hence by the FTA, we will be able to.

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Presentation on theme: "The Euler Phi-Function Is Multiplicative (3/3) Last time we showed that if p is prime, then  (p k ) = p k – p k-1. Hence by the FTA, we will be able to."— Presentation transcript:

1 The Euler Phi-Function Is Multiplicative (3/3) Last time we showed that if p is prime, then  (p k ) = p k – p k-1. Hence by the FTA, we will be able to compute  (m) for any m provided that  is multiplicative (recall that definition!). To this end, let m and n be relatively prime. We define two sets: Let S = {a : 1 ≤ a ≤ mn and GCD(a, mn) = 1}. How elements does S have? Let T = {(b, c) : 1 ≤ b ≤ m and GCD(b, m) = 1, and 1 ≤ c ≤ n and GCD(c, n) = 1}. How many elements does T have?

2 Do S and T have the same number of elements? We will know then that  (mn) =  (m)  (n) if we can show that S and T have the same number of elements. We can do that by displaying a one-to-one and onto correspondence between the two sets. (This is basic set theory.) Okay, so let f be the function from S to T given by f (a) = (a (mod m), a (mod n)). Example: Suppose m = 8, n = 7, and a = 25. Then f (a) = (?, ?). We claim f is one-to-one and onto.

3 The function f is one-to-one We show “one-to-oneness” of a function by assuming that there are two elements of the domain which are sent to the same element of the range, and show that they were in fact the same element to start with. So assume a 1 and a 2 are in S, and suppose f (a 1 ) = f (a 2 ), i.e., a 1 (mod m) = a 2 (mod m) and a 1 (mod n) = a 2 (mod n). Hence m | (a 2 – a 1 ) and n | (a 2 – a 1 ), but then, by Exercise 7.2 (which you did!), we get that …….. Thus a 1 = a 2 (why?), and so we have established that f is one-to-one.

4 The function f is onto This is trickier. In fact this result is itself an important one in number theory and hence has its own name: The Chinese Remainder Theorem. If GCD(m, n) = 1, then the simultaneous congruences x  b (mod m) and x  c (mod n) have a unique solution x 1 with 0 ≤ x 1 < mn. This is best seen via an example. Suppose again that m = 8 and n = 7. What is a solution less than 56 to to the congruences x  1 (mod 8) and x  4 (mod 7)? Well, we already saw that 25 is a solution. This theorem (CRT) says it will be only solution as well.

5 An algorithm to solve simultaneous congruences Consider again x  1 (mod 8) and x  4 (mod 7). What’s x (below 56)? Since x  1 (mod 8), we know there exists a y such that x = 1 + 8y. Plug this into the second congruence: 1 + 8y  4 (mod 7), so 8y  3 (mod 7). But we know that since GCD(7, 8) = 1, this linear congruence has a unique solution! Using our EEA algorithm to solve it (or by reducing the 8 (mod 7)!), we get y = 3. Hence x = 1 + 8(3) = 25, as predicted! Why is 25 unique? Well, y was unique below 7, but then x had to be unique and at most 7 + 8(6) = 55.

6 Final comments on CRT The CRT can easily be extended to any collection of moduli (not just 2 of them), provided that they are pairwise relatively prime. Please read the general proof of CRT in the text (pages 79-80) and please do Exercises 11.5 and By the way, please note that in the course of all this, we have proved that the Euler phi-function is indeed multiplicative. Yea! Hence, if we can factor m, we can compute  (m). For large m, this turns out to be a “big if”.


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