Find the smallest Pythagorean triangle, satisfying the following conditions:

1. its perimeter is a perfect square.
2. its area is a perfect cube.

(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 Multi-Primitive 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 non-commutative 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 2014-03-11 09:14:57