(In reply to

re: Computer exploration by Jer)

please edit

(3^n)^3 + (2*3^n)^3 = (1+2^3)*(3^n) = 9*(3^n)= 3^(n+2)

So if the coprime requirement is lifted - your answer would become

The (**3m+2)** powers of 3 can be written as the sum of two 3rd powers of integers (with a common factor of 3^3m):

**3^(3m+2)**=**(**3^(3m))(1+8)= **3^(3m)** + **(2*3^m)^3**

**e.g. for m=1 3^5= 3^3+6^3= 27+216=243**

*Edited on ***August 26, 2016, 8:38 am**