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 Subject: "pythagorean triples" Previous Topic | Next Topic
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h20skiier4life guest
Mar-20-02, 09:48 PM (EST)

"pythagorean triples"

 i have a hw problem for my geom. class. it'states:1)list the pythagorean triples generated using <= 52)Find expressions for m, and n in terms of a, b, and c. 3)If you are given 3 numbers and asked to find the corresponding values of m and n, how can you decide which number is a, b, and c?4)find values of m and n for the pythagorean triple 56, 90, 106.5)find values of m and n for pythagorean triple 48, 55, 73.i have no clue where to even start. if anyone can help me, it would be greatly appreciated : whiteknite13@hotmail.coma = n^2 - m^2b = 2mnc = n^2 + m^2

Member since Jun-22-03
Sep-05-03, 08:40 PM (EST)    1. "RE: pythagorean triples"
In response to message #0

 LAST EDITED ON Sep-06-03 AT 02:30 AM (EST) List the pythagorean triples generated using <= 5.a = n2 - m2b = 2�n�mc = n2 + m2Since a > 0, it must be n > m. Since b is even, even a at the same time would lead to a Pythagorean triple that is a multiple of some lower Pythagorean triple and we will skip those. Consequently, n and m cannot be both even or both odd.n = 2, m = 1a = 22 - 12 = 4 - 1 = 3b = 2�2�1 = 4c = 22 + 12 = 4 + 1 = 5n = 3, m = 2a = 32 - 22 = 9 - 4 = 5b = 2�3�2 = 12c = 32 + 22 = 9 + 4 = 13n = 4, m = 1a = 42 - 12 = 16 - 1 = 15b = 2�4�1 = 8c = 42 + 12 = 16 + 1 = 17n = 4, m = 3a = 42 - 32 = 16 - 9 = 7b = 2�4�3 = 24c = 42 + 32 = 16 + 9 = 25n = 5, m = 2a = 52 - 22 = 25 - 4 = 21b = 2�5�2 = 20c = 52 + 22 = 25 + 4 = 29n = 5, m = 4a = 52 - 42 = 25 - 16 = 9b = 2�5�4 = 40c = 52 + 42 = 25 + 16 = 41---------------------------------------------------------------------Find expressions for m, and n in terms of a, b, and c.c + a = 2n2c - a = 2m2n = �{(c + a)/2}m = �{(c - a)/2}Alternately:c + b = (n + m)2c - b = (n - m)2�(c + b) = n + m�(c - b) = n - mn = 1/2�{�(c + b) + �(c - b)}m = 1/2�{�(c + b) - �(c - b)}---------------------------------------------------------------------If you are given 3 numbers and asked to find the corresponding values of m and n, how can you decide which number is a, b, and c?Since c > a and c > b, c is the largest integer of the triple.Since (c + a)/2, (c - a)/2, c + b , and c - b are all complete squares:(c + a)/2 = n2(c - a)/2 = m2c + b = (n + m)2c - b = (n - m)2where n and m integers, neither of c + a, c - a, (c + b)/2, or (c - b)/2 can be a complete square:c + a = 2n2c - a = 2m2(c + b)/2 = (n + m)2/2(c - b)/2 = (n - m)2/2Try to add both the remaining numbers to c and/or to subtract both the remaining numbers from c. b is the one that makes a complete square when added to or subtracted from c.---------------------------------------------------------------------Find values of m and n for the Pythagorean triple 56, 90, 106.Since 106 > 56 and 106 > 90, c = 106106 + 56 = 162 is not a complete square106 - 56 = 50 is not a complete square106 + 90 = 196 = 142 is a complete square106 - 90 = 16 = 42 is a complete squarea = 56b = 90n = �{(c + a)/2} = �{(106 + 56)/2} = �(162/2) = �81 = 9m = �{(c - a)/2} = �{(106 - 56)/2} = �(50/2) = �25 = 5Alternatelyn = 1/2�{�(c + b) + �(c - b)} = 1/2�{�(106 + 90) + �(106 - 90)} == 1/2�(�196 + �16) = 1/2�(14 + 4) = 18/2 = 9m = 1/2�{�(c + b) - �(c - b)} = 1/2�{�(106 + 90) - �(106 - 90)} == 1/2�(�196 - �16) = 1/2�(14 - 4) = 10/2 = 5---------------------------------------------------------------------Find values of m and n for the Pythagorean triple 48, 55, 73.Since 73 > 48 and 73 > 55, c = 73 73 + 48 = 121 = 112 is a complete square73 - 48 = 25 = 52 is a complete square73 + 55 = 128 is not a complete square73 - 55 = 18 is not a complete squarea = 55 b = 48n = �{(c + a)/2} = �{(73 + 55)/2} = �(128/2) = �64 = 8m = �{(c - a)/2} = �{(73 - 55)/2} = �(18/2) = �9 = 3Alternatelyn = 1/2�{�(c + b) + �(c - b)} = 1/2�{�(73 + 48) + �(73 - 48)} == 1/2�(�121 + �25) = 1/2�(11 + 5) = 16/2 = 8m = 1/2�{�(c + b) - �(c - b)} = 1/2�{�(73 + 48) - �(73 - 48)} == 1/2�(�121 - �25) = 1/2�(11 - 5) = 6/2 = 3

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