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• ## 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
Dec 1 3:21 PM 1 of 9
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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
Dec 2 8:37 PM 1 of 9
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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
• ... Cheers, David ... From: Marcel Martin Cc: Sent: Sunday, December 02, 2001 10:16 PM Subject: Re:
Dec 3 3:48 AM 1 of 9
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With you now, cheers. Does this explain the other two p. triangles:

> 21 -> 28 -> 35
> 21 -> 72 -> 75

> 15 -> 20 -> 25
> 15 -> 36 -> 39

> 33 -> 44 -> 55
> 33 -> 180 -> 183

> 35 -> 84 -> 91
> 35 -> 120 -> 125

> These dimensions are for such p. triangles that when you subtract
> the two dervied lengths from given number they produce the factors
> e.g.
> 77 -> 264 -> 275
> 77 -> 420 -> 427

> 275 - 264 = 11
> 427 - 420 = 7

> 7 * 11 = 77

Cheers,
David

----- Original Message -----
From: "Marcel Martin" <znz@...>
Sent: Sunday, December 02, 2001 10:16 PM
Subject: Re: [PrimeNumbers] Pythagoras and factoring

>
> I sent an other post in which I pointed out the typo I made in the
> first one. The right relation was
> 29^2 - 21^2 = (5^2 + 2^2)^2 - (5^2 - 2^2)^2.
>
>
> Fermat: N = u^2 - v^2
> Yours : N^2 = U^2 - V^2
>
> Assuming N = a*b, a and b odd, a > b.
> With Fermat, it comes, u = (a+b)/2 and v = (a-b)/2. Then your
> parameters are simply U = u^2 + v^2 and V = 2uv
>
> With 35 you get
> u = 6, v = 1 thus U = 37 and V = 12.
>
> >and so
> >37^2 - 12^2 != (7^2 + 5^2)^2 - (7^2 - 5^2)^2
>
> Of course. But
>
> 37^2 - 35^2 = (6^2 + 1^2)^2 - (6^2 - 1^2)^2
> U^2 - N^2 = (u^2 + v^2)^2 - (u^2 - v^2)^2
>
> >Will look deeper but perhaps the only relation to Fermat's method is that
it
> >uses squares.
>
> No. It is just a complication of Fermat method. And, except maybe
> some particular cases, it is much less efficient since you have to
> work with numbers that are twice greater than the ones required in
> the Fermat method.
>
> Marcel Martin
>
>
> Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
> The Prime Pages : http://www.primepages.org
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
>
>
>
• ... 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
Dec 3 4:07 AM 1 of 9
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> 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
Dec 3 3:23 PM 1 of 9
View Source
> 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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