- 25 Dec '08 13:23 / 1 editUsing 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.
- 25 Dec '08 13:31

There might be at least ten of them :-*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.**

(3,4,5),

(5,12,13),

(8,15,17),

(20,21,29), ....... etc..etc... - 27 Dec '08 09:02

You forget. the condition of independent triplets.*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),...

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. - 27 Dec '08 14:02 / 1 edit

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

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. - 28 Dec '08 14:59
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. - 28 Dec '08 23:39 / 1 edit

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

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. - 29 Dec '08 02:55

I have been able to discover at least 2 more than Geepamoogle...*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.

12-35-37 and 27-36-45...Hence there are at least 18. May be more..