Pythagorean triplets

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...

Originally posted by howzzat
There might be at least ten of them :-

(3,4,5),
(5,12,13),
(8,15,17),
(20,21,29), ....... etc..etc...

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),...

I found 16 of them, and I strongly believe there are only 16 unique triplets (excluding 0,1,1)

Good exercise is to discover them all..

I'll post the hypotenuses of them all in a day or two.

I could only find 15 of them and thought I had done it by exhaustion (would love to know a better way). Cant see the one I have missed if there are 16

**Edit
I cant count I agree there are 16!

There is a known series of formulas for Pythagorean triples. That's what I used.

Also important to note that you only need worry that the hypotenuse be less than 100.

If you didn't care about ratios of sides being distinct, I count 57 triples.

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.

Originally posted by geepamoogle
I found 16 of them, and I strongly believe there are only 16 unique triplets (excluding 0,1,1)

Good exercise is to discover them all..

I'll post the hypotenuses of them all in a day or two.

Plz give the list of all the 16 triplets discovered by you.

I thought there were infinite number of pythagorean triangles?

Originally posted by ranjan sinha
Plz give the list of all the 16 triplets discovered by you.

Very well.

3-4-5
5-12-13
8-15-17
7-24-25
20-21-29
12-35-37
9-40-41
28-45-53
11-60-61
16-63-65
33-56-65
48-55-73
13-84-85
36-77-85
39-80-89
65-72-97

The triplets formulas which help find these quickly and thoroughly are as follows (where a>b).

2*a*b
a^2 - b^2
a^2 + b^2 < hypotenuse

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

Originally posted by FabianFnas
I thought there were infinite number of pythagorean triangles?

There probably is, but the thread author wants hypotenuse's less than 100

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.

Originally posted by geepamoogle
Very well.

3-4-5
5-12-13
8-15-17
7-24-25
20-21-29
12-35-37
9-40-41
28-45-53
11-60-61
16-63-65
33-56-65
48-55-73
13-84-85
36-77-85
39-80-89
65-72-97

The triplets formulas which help find these quickly and thoroughly are as follows (where a>b).

2*a*b
a^2 - b^2
a^2 + b^2 < hypotenuse

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

I have been able to discover at least 2 more than Geepamoogle...
12-35-37 and 27-36-45...Hence there are at least 18. May be more..