I cannot tell the answer to your question (I guess: NO), however you have tagged the question with C++ , so I give you my simple implementation in C/C++ to lookup for such triples.

I've run it for n=1000000 (44 minutes), that means a

Dear Romeo Meštrović ,

Solve for a, b, c, d :

a² + b² = c²....(1)

a³ + b³ + c³ = d³.....(2)

The general solution for (1) is well known:

a = s² - t² , b = 2st, c = s²+t² where gcd(s,t)=1.

Now, substitute in(2), we obtain :

(s²-t²)³ + (2st)³+ (s²+t²)³ - d³ = 0,

simplify: 2s²(s²-2st+3t²)(s+t)² = d³

hence, d is an even number, let d= 2kD , k > 0, and D is odd, we get

s²(s²-2st+3t²)(s+t)² = 23k-1D³ ......(3)

Consider the 3 accepted cases:

(I) s = 2m+1, t = 2n, we have,

(4m-4n-8mn+4m²+12n²+1)(2m+2n+1)² (2m+1)² = 23k-1D³ contradiction.

(II) s = 2m+1, t = 2n+1

8(4n-4mn+2m²+6n²+1)(m+n+1)²(2m+1)² =23k-1D³

implies 23k-1 = 8 = 23 , therefore 3k-1= 3 contradiction.

The only possible case is :

(III) s = 2m, t = 2n+1,

4(12n-4m-8mn+4m²+12n²+3)(2m+2n+1) m²=23k-1D³

hence, k=1 and m should be odd, m = 2r+1 and d=2D with odd D.

The work is in progress. To handle the last case.

Best regards

Dear Tibor Tajti and Issam Kaddoura,

Thank you very much for interesting on my question/problem. I have also applied the same approach as in Kaddourra previous answer to the eventual solving of the problem. However, it seems that this is very difficult Number Theory/Diophantine Equation problem. Unfortunately, a computation by Tibor Tajti C/C++ did not reveal any new one irreducible super Pythagorean triple. Moreover, I believe that some theoretical results would be (at least) contribute to a faster computational search of irreducible super Pythagorean triples (such as well known formulae for Pythagorean triples – attributed in the previous Kaddoura’s answer, which are not used in the implementation of the problem in C/C++ presented by Tibor Tajti). I have implemented these formulae in a command in Wolfram's Mathematica 9 for searching irreducible super Pythagorean triples of the form {m^2-n^2},{2*m*n},{m^2+n^2} with m > n, but until this moment no new triples different from (3, 4, 5) has been obtained for m < 10000.

Best regards