We began with the solution to the preceding problem, but with slightly different notation. We had called certain sets of integers "threven," "throdd_1", and "throdd_2" in analogy with "even" and "odd." But more conventional names for them would be just
"0"={0,3,6,9,12,...}
"1"={1,4,7,10,13,...}
"2"={2,5,8,11,14,...}
In what follows we will often leave out the " ", although it is clear that "2" above is NOT the integer 2. Rather it is a big set, vaguely analogous to the "odd numbers." There is an arithmetic on these sets, analogous to the familiar rules for adding and multiplying even and odd numbers, which was essentially today's problem:
| + | "0" | "1" | "2" |
| "0" | "0" | "1" | "2" |
| "1" | "1" | "2" | "0" |
| "2" | "2" | "0" | "1" |
| * | "0" | "1" | "2" |
| "0" | "0" | "0" | "0" |
| "1" | "0" | "1" | "2" |
| "2" | "0" | "2" | "1" |
It is interesting to note that "0" really behaves like 0: when you add it, it doesn't change anything, and when you multiply anything by it you get "0" as the result. Thus 0 would really be a good name for it, even without the " ". Also "1" really behaves like 1, pretty much. When you add it, the result goes up by one (except for cycling around from "2" back to "0"), and when you multiply anything by "1", it doesn't change. Another name for "2", maybe a better name, would be "-1." Adding "2" decreases "2" to "1" and "1" to "0". That seems like adding -1. Also (-1)*(-1)=1 looks more intuitive than 2*2=1, which is what we have in the lower right corner of the * table. But it is a matter of taste what we call these sets. The class actually voted for "2" rather than "-1", although the vote was not overwhelming.
Here is an application of arithmetic mod 3 to primitive Pythagorean triples. Recall that these are solutions in integers (x,y,z) to the Diophantine equation
The solution is called a primitive Pythagorean triple if (x,y,z) have no factors in common. For example, (3,4,5) is a primitive Pythagorean triple, while (6,8,10) is a Pythagorean triple, but not primitive, because the factor 2 is common to all three numbers in the triple. We know other examples: (5,12,13), (8,15,17), (9,40,41), (11,60,61), (7,24,25), etc. Each of these, if you replace by its representative in mod 3 arithmetic, leads to a solution of the above equation mod 3. For example, since 3^2=9=0 mod 3, 4^2=16=1 mod 3, and 5^2=25=1 mod 3, we find the primitive triple (3,4,5) becomes
which is certainly a true relationship. In fact, what solutions ARE there, mod 3? If we look at the multiplication table, we see that the only squares mod 3 are 0 and 1. The only ways to solve the equation with 0's and 1's mod 3 are
The last one, however, couldn't come from a primitive triple, because (x,y,z) would have the factor 3 in common. Thus only the first two actually occur in primitive Pythagorean triples, and in particular the 0's mean that exactly one of (x,y) must be a multiple of 3! We see in the examples that this is true, but it is apparently true for all primitive triples -- not obvious, but true.
Fill in the multiplication and addition tables for arithmetic mod 5. This is arithmetic on the sets
"0"={0,5,10,15,20,25,...}
"1"={1,6,11,16,21,26,...}
"2"={2,7,12,17,22,27,...}
"3"={3,8,13,18,23,28,...}
"4"={4,9,14,19,24,29,...}
in analogy to "even and odd" and arithmetic mod 3. For example
Fill in all of both tables, both addition and multiplication.
There is an optional part, more interesting! What can arithmetic mod 5 tell us about primitive Pythagorean triples? A look at the examples suggests that there is always exactly one multiple of 5. Is this true? Try to prove it!