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   Integer triangle with 120 degrees
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   Author  Topic: Integer triangle with 120 degrees  (Read 2875 times)
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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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.
« Last Edit: Apr 23rd, 2013, 10:43am by towr » IP Logged

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Christine
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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 ?
« Last Edit: Apr 23rd, 2013, 12:04pm by Christine » IP Logged
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