Originally posted by ranjan sinhaThere might be at least ten of them :-
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 cosmic voiceYou forget. the condition of independent triplets.
There are more than 10 of them sure enough -
Originally posted by ranjan sinhaVery well.
Plz give the list of all the 16 triplets discovered by you.
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.
Originally posted by afxIf both are odd, then all three terms end up being even.
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 geepamoogleI have been able to discover at least 2 more than Geepamoogle...
The triplets formulas which help find these quickly and thoroughly are as follows (where 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.