## Re: a general formula for the following

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• ... Yes: (4*n^2+1)^2 - (4*n^2-1)^2 = (4*n)^2 ... No: for p = 0 mod 4, the number of odd coprime pairs [x,y] with p = x^2 - y^2 is 2^k where k is the number of
Message 1 of 3 , Apr 2, 2009
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"san_tan1" <san_tan1@...> wrote:

> 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

Yes: (4*n^2+1)^2 - (4*n^2-1)^2 = (4*n)^2

> 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)....up
> to any desired no. of duplets such that in each duplet(x,y)
> x,y are both odd and mutually prime

No: for p = 0 mod 4, the number of odd coprime pairs [x,y]
with p = x^2 - y^2 is 2^k where k is the number of distinct
odd prime divisors of p. This procedure will print them all:

{pairs(p)=local(x,y,c,d);if(p%4==0,fordiv(p/2,d,
c=p/2/d;if(c>d,x=c^2+d^2;y=c^2-d^2;if(gcd(x,y)==1,
print([x,y])))));}

pairs(2^3*3^2*5*7*11);

[192099601, 192099599]
[12006241, 12006209]
[7684009, 7683959]
[3920449, 3920351]
[2371681, 2371519]
[1587721, 1587479]
[480649, 479849]
[245809, 244241]
[158041, 155591]
[149521, 146929]
[101161, 97289]
[96889, 92839]
[66529, 60479]
[52369, 44431]
[38329, 26471]
[29401, 9799]

David
• PS: I left out a square sign, here restored: for p = 0 mod 4, the number of odd coprime pairs [x,y] with p^2 = x^2 - y^2 is 2^k where k is the number of
Message 2 of 3 , Apr 2, 2009
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PS: I left out a square sign, here restored:

for p = 0 mod 4, the number of odd coprime pairs [x,y]
with p^2 = x^2 - y^2 is 2^k where k is the number of distinct
odd prime divisors of p.

David^2
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