*j*and

*k*and let

*j*be the larger. Here are the formulas for

*a, b*and

*c*.

*a*=

*j*² -

*k*²

*b*= 2

*jk*

*c*=

*j*² +

*k*²

**Example #1:**

*j*= 2,

*k*= 1

*a*= 4-1 = 3

*b*= 2 × 2 × 1 = 4

*c*= 4+1 = 5

**Example #2:**

*j*= 3,

*k*= 1

*a*= 9-1 = 8

*b*= 2 × 3 × 1 = 6

*c*= 9+1 = 10

**Example #3:**

*j*= 3,

*k*=2

*a*= 9-4 = 5

*b*= 2 × 3 × 2 =12

*c*= 9+4 = 13

Not surprisingly, using the three smallest pairs around (2, 1) (3, 1) and (3, 2), we get the three smallest Pythagorean triples. But notice that (3, 1) generates 8-6-10 which is

*not*a relatively prime triple, since all the numbers are even. Looking at this carefully, we see that if both

*j*and

*k*are odd, this will make

*a, b*and

*c*even.

*a*= odd² - odd² = even

*b*= 2×odd×odd = even

*c*= odd² + odd² = even

To generate a relatively prime Pythagorean triple, we need two things to be true about

*j*and

*k*.

1.

*j*and

*k*must be relatively prime.

2.

*j - k*must be odd. (Another way to state this is one is odd and the other even.)

Tomorrow, a few more rules about the relatively prime Pythagorean triples.

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