wu :: forums « wu :: forums - Right triangles and semiprimes » Welcome, Guest. Please Login or Register. Sep 11th, 2024, 8:53pm RIDDLES SITE WRITE MATH! Home Help Search Members Login Register
 wu :: forums    riddles    medium (Moderators: ThudnBlunder, towr, Icarus, SMQ, Eigenray, Grimbal, william wu)    Right triangles and semiprimes « Previous topic | Next topic »
 Pages: 1 Reply Notify of replies Send Topic Print
 Author Topic: Right triangles and semiprimes  (Read 909 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

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 »