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Topic: Integer triangle with 120 degrees (Read 2889 times) |
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Christine
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Integer triangle with 120 degrees
« on: Apr 23rd, 2013, 10:35am » |
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This type of triangle can be generated by a = m2 - n2 b = 2mn + n2 c = m2 + mn + n2 we note a2 + ab + b2 = c2 If m = 2, n = 1 ---> a = 3, b = 5, c = 7 the smallest triangle with an angle of 120 degrees having all 3 sides with prime numbers. Is it the only triangle with the three sides being prime numbers? If yes, how to prove it?
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towr
wu::riddles Moderator Uberpuzzler
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Re: Integer triangle with 120 degrees
« Reply #1 on: Apr 23rd, 2013, 10:41am » |
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Well, n has to be 1, otherwise b can't be prime (n divides 2mn + n^2). And m-n has to be 1, because otherwise a isn't prime (m-n divides m^2-n^2). So therefore m=2, n=1 is the only solution with prime numbers. Provided all such triangles can be generated in this way.
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« Last Edit: Apr 23rd, 2013, 10:43am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Christine
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Posts: 159
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Re: Integer triangle with 120 degrees
« Reply #2 on: Apr 23rd, 2013, 11:13am » |
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on Apr 23rd, 2013, 10:41am, towr wrote:Well, n has to be 1, otherwise b can't be prime (n divides 2mn + n^2). And m-n has to be 1, because otherwise a isn't prime (m-n divides m^2-n^2). So therefore m=2, n=1 is the only solution with prime numbers. Provided all such triangles can be generated in this way. |
| Can n be a prime number > 2 ?
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« Last Edit: Apr 23rd, 2013, 12:04pm by Christine » |
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