The Euler Phi-Function Is Multiplicative
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• Last time we showed that if p is prime, then
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(pk) = pk – pk-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?
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.
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 a1 and a2 are in S, and suppose f (a1) = f (a2),
i.e., a1 (mod m) = a2 (mod m) and a1 (mod n) = a2 (mod n).
• Hence m | (a2 – a1) and n | (a2 – a1), but then, by Exercise
7.2 (which you did!), we get that ……..
• Thus a1 = a2 (why?), and so we have established that f is
one-to-one.
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 x1 with 0 ≤ x1 < 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.
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.
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 11.6.
• 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”.