(In reply to
The 21 smallest by Charlie)
Yes.
There is a solution (eventually) for every primitive triple (and all triples).
Take the primitive triple {a,b,h}, and compute P=(a+b+h) and A=(1/2ab). Determine the prime factors of both numbers. Next determine the MultiPrimitive Value (MPV) as follows:
1. Each factor that occurs to the same power in both P and A will occur to exactly the same power in MPV.
2. Each factor that occurs to different powers in P and A will occur in MPV to the power (p+x) where p+x=2y, a+2x=3z, with p and a the powers of that factor in P and A respectively, leading to a noncommutative table (p+a gives x) beginning:
0+0 gives 0 0+1 gives 4 0+2 gives 2 0+3 gives 6
1+0 gives 3 1+1 gives 1 1+2 gives 5 1+3 gives 3
2+0 gives 6 2+1 gives 4 2+2 gives 2 2+3 gives 6
3+0 gives 3 3+1 gives 1 3+2 gives 5 3+3 gives 3
3. Multiply each of a,b,h by the MPV. The result will be some triple MPV*{a,b,h}, for which the stipulation of the problem holds good.
4. In like manner, it is easily shown that the smallest such triangle is 48*{3,4,5}.
Edited on March 11, 2014, 9:55 am

Posted by broll
on 20140311 09:14:57 