Using the natural numbers less than 100 only, howmany distinct (independent) pythagorean triplets can you form? Remember (0.1,1) is not allowed. Also (3,4,5) and (6,8,10) are not to be treated independent triplets. In other words only the triplets in their lowest multiple form, if distinct, are to be counted as distinct and independent triplets.

Originally posted by ranjan sinha Using the natural numbers less than 100 only, howmany distinct (independent) pythagorean triplets can you form? Remember (0.1,1) is not allowed. Also (3,4,5) and (6,8,10) are not to be treated independent triplets. In other words only the triplets in their lowest multiple form, if distinct, are to be counted as distinct and independent triplets.

There might be at least ten of them :-
(3,4,5),
(5,12,13),
(8,15,17),
(20,21,29), ....... etc..etc...

I could have used the formulas but an excel spreadsheet got me the answer quicker. It is a bit lazy of me but there again it is Christmas. Only when I use a less elegant way I feel a bit guilty about out.

Originally posted by cosmic voice There are more than 10 of them sure enough -
(7,24,25),
(9,40,41),
(33, 56,65),
(16, 63,65),
(10,24,26),
(12,35,37),...

You forget. the condition of independent triplets.
Actually . (10,24,26) is not an independent triplet. It is the same as (5,12,13)..And hence it is not to be counted as sepatete triplet.

If a and b are relatively prime AND exactly one of the two is even, you'll find a unique triplet.

I needed some time to understand, why EXACTLY ONE of the two must be even.
But finally I got it:
When a and b both are even, they are not relatively prime.
Let (2n+1) and (2m+1) be two odd numbers, then
(2n+1)^2 + (2m+1)^2 = 4( n^2+m^2 + n+m) + 2
so the sum of the squares mod 4 is 2 and can therefore be no square number.

Originally posted by afx I needed some time to understand, why EXACTLY ONE of the two must be even.
But finally I got it:
When a and b both are even, they are not relatively prime.
Let (2n+1) and (2m+1) be two odd numbers, then
(2n+1)^2 + (2m+1)^2 = 4( n^2+m^2 + n+m) + 2
so the sum of the squares mod 4 is 2 and can therefore be no square number.

If both are odd, then all three terms end up being even.

Take a=3, b=1 for example.

2*3*1 = 6
3^2 - 1^2 = 8
3^2 + 1^2 = 10

But 6-8-10 is just 3-4-5 doubled.

With either a or b even, though, two of the terms are odd, including the hypotenuse.