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Topic: Right triangles and semiprimes (Read 895 times) 

Christine
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Right triangles and semiprimes
« on: Aug 1^{st}, 2014, 10:31am » 
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I looked at the number of different right triangle with a leg equal to a semiprime less than 100 I found that when the legs are given the values 15,21,33,35,51,55,65,77,85,87,91,93,95 I get exactly four right triangles. e.g. 15,112,113),(15,20,25),(15,36,39),(8,15,17) Why do we get four right triangles with these semiprimes?


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towr
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Re: Right triangles and semiprimes
« Reply #1 on: Aug 1^{st}, 2014, 1:12pm » 
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I can explain at least 3, because you can get at least one right triangle for every odd n. And since an odd semiprime has two odd factors and is itself odd, that gives three. Any primitive Pythagorean triple can be written as a = m^{2}  n^{2}, b = 2mn, c = m^{2} + n^{2} for m>n coprime and mn odd. So with m=n+1: a = 2n+1, b= [ (2n+1)^{2}  1]/2, c = [ (2n+1)^{2} + 1]/2 So for an odd semiprime s = p*q we have at least the following square triangles with side s q*p, q*[p^{2}  1]/2, q*[p^{2} + 1]/2 p*q, p*[q^{2}  1]/2, p*[q^{2} + 1]/2 p*q, [p^{2}q^{2}  1]/2, [p^{2}q^{2} + 1]/2 [edit] A fourth triple comes from m,n = (p+q)/2, (pq)/2 p*q, [p^{2}  q^{2}]/2, [p^{2} + q^{2}]/2 However, it's still possible that for bigger semiprimes there are more triples than these 4. [/edit]

« Last Edit: Aug 1^{st}, 2014, 1:23pm by towr » 
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towr
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Re: Right triangles and semiprimes
« Reply #2 on: Aug 1^{st}, 2014, 2:19pm » 
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To get m^{2}  n^{2} = p*q, you need p*q = 2nx + x^{2} (for m = n+x) Since p<q are prime, either the p is x and n = (qp)/2, or x=1 and n=(p*q1)/2 So that gives two primitive triples with p*q as odd leg And 2 with p and q as odd leg, for a total of 4.


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