7-24-25 , 33-56-65 to name a couple.
There is a general algorithm to find them:
take a and b to be positive integers, a < b.
Then (a^2 - b^2)^2 + (2ab)^2 = a^4 -2a^2b^2 + b^4 +4a^2b^2 = a^4 +2a^2b^2 + b^4 = (a^2 + b^2)^2
so [a^2 - b^2]-[2ab]-[a^2 + b^2] is always a pythagorean triple if a and b are integers, with a < b.
2007-03-16 18:29:25
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answer #1
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answered by mitch w 2
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There are many. In fact, there are 16 where the third number is less than 100:
(3, 4, 5) (20, 21, 29) (11, 60, 61) (13, 84, 85)
(5, 12, 13) (12, 35, 37) (16, 63, 65) (36, 77, 85)
(8, 15, 17) (9, 40, 41) (33, 56, 65) (39, 80, 89)
(7, 24, 25) (28, 45, 53) (48, 55, 73) (65, 72, 97)
A simple recipe for generating more is at the link I provided.
2007-03-17 01:31:28
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answer #2
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answered by Sam 5
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Of course any multiple of these will be, such as 6-8-10, 9-12-15 etc, but I assume that is not what you want.
7-24-25
9-40-41
11-60-61
13-84-85
15-112-113
12-35-37
16-63-65
20-99-101
24-143-145
28-195-197
How's that for a start?
2007-03-17 01:33:47
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answer #3
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answered by yupchagee 7
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this method generally works
take and odd number and
square it and divide it by
2. Say 7, its square is 49
it can be split into two parts
24 and 25 , they form triplets
7-24-25,9(81)-40-41,11-60-61 etc.
2007-03-17 01:32:34
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answer #4
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answered by more1708_par 2
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U can compute them using
a=m^2+n^2; b=m^2-n^2; c=2mn where m and n are integers and m>n.
2007-03-17 01:46:43
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answer #5
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answered by Ahmad k 2
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