## x**2 - y**2 = p**2

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• 5. a general formula for the following Posted by: san_tan1 san_tan1@yahoo.co.in san_tan1 Date: Thu Apr 2, 2009 5:15 am ((PDT)) is there any general formula
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5. a general formula for the following
Posted by: "san_tan1" san_tan1@... san_tan1
Date: Thu Apr 2, 2009 5:15 am ((PDT))

is there any general formula for triples (a,b,c) such that a,b are
mutually prime and both odd and also

a**2-b**2=c**2. ....(1)?

also can this be extended to finding an algorithm for generating duplets
(a,b) (c,d) etc....such that given p we can find upto any number of
desired duplets having property 1??

that is given p,

p**2= a**2-b**2=c**2-d**2=e**2-f**2.......
is there any way to generate (a,b) (c,d) ,(e,f)....upto any desired no.
of duplets such that in each dupet(x,y) x,y are both odd and mutually
prime..?

Hello san_tan1.

First we strive to make the formulas without regard to whether or
not x and y are prime.

if
x = m1 * m2 + n1 * n2
and
y = m1 * n2 - m2 * n1

then x**2 - y**2
= (m1**2 m2**2 + n1**2 n2**2 - m1**2 n2**2 - m2**2 n1**2 )
= m1**2 (m2**2 - n2**2) + n1**2 (n2**2 - m2**2)
= m1**2 (m2**2 - n2**2) - n1**2 (m2**2 - n1**2)
= (m1**2 - n1**2) (m2**2 - n2**2)

= (m1 - n1)(m1+n1)(m2-n2)(m2+n2)

p**2 = (m1 - n1)(m1+n1)(m2-n2)(m2+n2)

m1 - n1 = (q1**2)*q2*q3*q4
m1 + n1 = (q5**2)*q2*q6*q7
m2 - n2 = (q8**2)*q3*q6*q9
m2 + n2 = (q10**2)*q4*q7*q9

m1 = ((q5**2)*q2*q6*q7 + (q1**2)*q2*q3*q4 )/2

n1 = ((q5**2)*q2*q6*q7 - (q1**2)*q2*q3*q4 )/2

m2 = ( (q10**2)*q4*q7*q9 + (q8**2)*q3*q6*q9 )/2

n2 = ( (q10**2)*q4*q7*q9 - (q8**2)*q3*q6*q9 )/2

x = m1 * m2 + n1 * n2

y = m1 * n2 - m2 * n1

Systematically assign values to q1,q2,. . . q10,

and search for x and y being prime.

There are probably ways to shortcut the search,
but I don't know them.

Kermit
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