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wu :: forums    riddles    medium (Moderators: william wu, SMQ, Eigenray, ThudnBlunder, Grimbal, towr, Icarus)    Right triangles and semiprimes « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Right triangles and semiprimes  (Read 923 times) Christine Full Member     Posts: 159 Right triangles and semiprimes   « on: Aug 1st, 2014, 10:31am » Quote Modify 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? IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: Right triangles and semiprimes   « Reply #1 on: Aug 1st, 2014, 1:12pm » Quote Modify 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 = m2 - n2,  b = 2mn, c = m2 + n2 for m>n coprime and m-n 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*[p2 - 1]/2, q*[p2 + 1]/2 p*q, p*[q2 - 1]/2, p*[q2 + 1]/2 p*q, [p2q2 - 1]/2, [p2q2 + 1]/2   [edit] A fourth triple comes from m,n = (p+q)/2, (p-q)/2 p*q,  [p2 - q2]/2, [p2 + q2]/2   However, it's still possible that for bigger semiprimes there are more triples than these 4. [/edit] « Last Edit: Aug 1st, 2014, 1:23pm by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: Right triangles and semiprimes   « Reply #2 on: Aug 1st, 2014, 2:19pm » Quote Modify To get m2 - n2 = p*q, you need p*q = 2nx + x2 (for m = n+x) Since p<q are prime, either the p is x and n = (q-p)/2, or x=1 and n=(p*q-1)/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. IP Logged Wikipedia, Google, Mathworld, Integer sequence DB Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy => medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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