Let s just do some examples to get the feel of congruence arithmetic.


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1 Basic Congruence Arithmetic Let s just do some examples to get the feel of congruence arithmetic. Arithmetic Mod 7 Just write the multiplication table Written this way, it is easy to see that the basic properties of arithmetic still hold. We have commutivity, and the products by 3, say, run up as we would expect as 0, 3, 6, 9 = 2, 12 = 5, 15 = 5, 18 = 5, 21 = 0. We note also that in any given row or column each of the linear residues modulo 7 appears exactly once. This is what has to happen in a field, and it always happens mod primes. Now let s rearrange the multiplication table based on the primitive root 3. Recall that the powers of 3 form a cycle 3, 9 = 2, 6, 18 = 4, 12 = 5, 15 = 1. And we re going to get rid of the zero row and column You should verify that I have not made a mistake in this table. Looked at this way, the cyclic nature of multiplicative arithmetic mod 7 is clear. 1
2 We can also do the obvious thing with exponents modulo 6. We have chosen 3 as the primitive root here, but we could also choose 5 as the primitive root. This would produce a rearranged multiplication table: The reason that we can choose either 3 or 5 is that the exponents work modulo 6, (mod 7), and (mod 7). If we tried to do this with any other power, we would short cycle. For example, since (mod 7), if we take powers of 2, we only get three distinct elements and not six: we get 2 2 (3 2 ) (mod 7) and 2 3 (3 2 ) (mod 7). And since (mod 7), if we take powers of 6, we only get two distinct elements: we get 6 2 (3 3 ) (mod 7). Arithmetic Mod 15 Now contrast this arithmetic modulo the prime 7 with arithmetic modulo the composite number 15. We get a different kind of multiplication table. 2
3 Not so pretty. We could rearrange the rows and columns somewhat, but first let s just look at the two short cycles. We have two subgroups that show up If we remove these rows and columns from the multiplication table modulo 15, we get the following multiplication table for those linear residues that are prime to 15. 3
4 Now this looks a little better. At least there are no short cycles. We also notice that we have 1,2,4,8 as a subcycle. In fact, with a little experimentation (and knowing in advance what we are trying to construct), we observe that the arithmetic of the eight linear residues modulo 15 that are relatively prime to 15 can be written out as The multiplicative arithmetic of the cycle of four elements {1, 2, 4, 8}, which is closed (i.e., all products of these elements come back as one of these elements. The multiplicative arithmetic of the two elements {1, 14}. which is closed, etc., just like the 4cycle of the previous bullet. Note that 14 1 (mod 15) and of course ( 1) 2 1 (mod anything). The multiplicative arithmetic of the three cross terms (mod 15), (mod 15), (mod 15). Note that we can factor out the arithmetic of the cross terms in the obvious way. For example, 11 7 (14 4) (14 8) (14 14) (2 4) (1) (8) (8) (mod 15) The way to view the arithmetic, in general, for moduli m that are products of two odd primes p and q (and this is mostly about as complicated as we will need to get) is this. Choose a primitive root a for p and form the cycle of powers of a, but take them modulo m and not modulo p. We just did this for the primitive root 2 modulo 5, but in this case we included the fourth power 8 as well as the powers 1, 2, and 4. 4
5 Choose a primitive root b for q. Among the elements b+q k for various k, choose one not in the cycle we just formed above that is prime to p. Mod 3, the primitive root is 2, and our choices for b are 2, 5, 8, 11, 14. We have already used 2 as the primitive root mod 5 and included 8 as a power of 2, so those two cannot be used. Since 5 is not prime to 5, we can t use it. This leaves 11 and 14, and we chose 14 and formed the cycle. Form the cross products. In general this method will work for any integer, not just products of two distinct primes. You should try this with 35. Arithmetic modulo powers of primes We have done arithmetic modulo primes above. This is easy, because the multiplication is a single cycle. The number of elements in the multiplicative cycle for an odd prime p is φ(p) = p 1. We have done arithmetic modulo products m = p q of two distinct odd primes. This is harder because we have to resolve the arithmetic of the φ(p) = p 1 elements modulo p and the φ(q) = q 1 elements modulo q and their cross terms. What we now need to look at is arithmetic modulo powers of primes. Arithmetic modulo powers of odd primes Arithmetic modulo 9 Consider the multiplication table modulo 9. We are, of course, going to throw out the residues 3 and 6, just as we got rid of the multiples of 3 and of 5 when working modulo 15. And we re going to arrange this in advance as a cycle. 5
6 The big theorem is that working modulo powers p k of an odd prime k, we still get a single cycle of the (p 1)p k 1 elements that are relatively prime to p. You should try this with 25. Arithmetic modulo powers of 2 The powers of 2 are different from the powers of any odd prime. Modulo 2, we get just binary (boolean) arithmetic with one nonzero element. Modulo 4, we have two elements Modulo 8 we have four elements in the multiplicative group, but they do not form a single cycle. Note that every element squares to This is what is called a Klein 4group. We saw that the arithmetic modulo 15 was formed from a 4cycle and a 2cycle and the cross products. The arithmetic modulo 8 is formed from a 2cycle (1 and 3, say) and a second 2cycle (1 and 5, say) and the cross product (mod 8). The arithmetic modulo powers of odd primes remains a single cycle. The arithmetic modulo higher powers of 2 is one 2cycle less than it should 6
7 be. Modulo 8, if powers of 2 followed the odd prime model, we d get a 4 cycle, but instead we get a product of two twocycles. Modulo 16, instead of an 8cycle, we get a product of a 2cycle and a 4cycle. Modulo 32, instead of a 16cycle, we get a product of a 2cycle and an 8cycle. And so forth. The arithmetic mod 16 is given below. Just as we generated the arithmetic modulo 15 from the powers of 2 and the powers of 14 and the cross terms, we can generate the arithmetic modulo 15 as the powers of 3 and the powers of 7 and the cross terms
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