## Re: [PrimeNumbers] Pythagoras and factoring

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• ... But: 205 - 164 = 41 123 / 41 = 3 So we can get the factors straight out with this p. triangle. David
Message 1 of 9 , Dec 1, 2001
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> Then B+C=(U+V)^2, B-C=(U-V)^2 as in your examples. But it is *not* common
> solution!
> For instance, the least solution to A^2+B^2=C^2 with A=123 is
>
> 123^2+164^2=205^2
>
> Both your C-program and my pascal one find this solution. But,alas
>
> 205-164=41 is *not* square, neither is
> 205+164=369=3^2*41.
>

But: 205 - 164 = 41
123 / 41 = 3
So we can get the factors straight out with this p. triangle.

David
• 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
Message 2 of 9 , Dec 2, 2001
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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:
>
> 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
• ... But: 205 - 164 = 41 123 / 41 = 3 So we can get the factors straight out with this p. triangle. David
Message 3 of 9 , Dec 3, 2001
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> Then B+C=(U+V)^2, B-C=(U-V)^2 as in your examples. But it is *not* common
> solution!
> For instance, the least solution to A^2+B^2=C^2 with A=123 is
>
> 123^2+164^2=205^2
>
> Both your C-program and my pascal one find this solution. But,alas
>
> 205-164=41 is *not* square, neither is
> 205+164=369=3^2*41.
>

But: 205 - 164 = 41
123 / 41 = 3
So we can get the factors straight out with this p. triangle.

David
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