Hello David,

I don't even know what the 'Fermat method' is, and I don't entirely

comprehend what has been exchanged thus far, but here are my

gleanings on the matter:

let a^2 + b^2 = c^2, a,b,c relatively prime. a or b must be even, so

let us assume b is odd.

b^2 = c^2 - a^2 = (c-a)(c+a)

It is clear that c-a and c+a are relatively prime, so each of c-a and

c+a must be squares themselves. So let

r^2 = c-a

s^2 = c+a

So then,

c = (s^2 + r^2)/2

a = (s^2 - r^2)/2

b = rs

If b is 21, the only combinations possible for r and s are

(r,s) = (1,21) and (3,7). That would make for two pythagorean

triplets easily calculated with the equations just above.

If b is 35, the only combinations possible for r and s are

(r,s) = (1,35) and (5,7). Two pythagorean triplets.

If b is 105 = 3*5*7, the combinations possible for r and s are

(r,s) = (3,35), (5,21), (7,15) and (1,105). Four pythagorean triplets.

More generally, if b is composed of x different prime factors, then

the number of pythagorean triples possible is 2^(x-1).

If b is a prime p, there is only one possible combination, and that is

(r,s) = (1,p).

In other words, if b is a prime p, the only pythagorean triplet

possible would be

b = p

a = (p^2 - 1)/2

c = (p^2 + 1)/2

Examples: (b,a,c), where a^2 + b^2 = c^2, b is prime:

(3,4,5), (5,12,13), (7,24,25), (11,60,61).

a little more mud in the water,

Mark

--- In primenumbers@y..., "David Litchfield" <Mnemonix@g...> wrote:

> > With you now, cheers. Does this explain the other two p.

triangles:

>

> Bad form answering your own emails (sorry) but I've just seen the

> relationship:

>

> Using 35 as the example

>

> 35 = 5 * 7

> p. triangle is 35 -> 120 -> 125

>

> 120 / 5 = 24

> 125 / 5 = 25

> 24 + 25 = 49

> 49 sqrt = 7

>

> and

>

> p. triangle is 35 -> 84 -> 91

> 84 / 7 = 12

> 91 / 7 = 13

> 12 + 13 = 25

> 25 sqrt = 5

>

>

> David