0,3

For n=2, there is an infinite number of solutions for the proposition. For n > 2, there is an infinite number of solutions to a^(1/3) + b^(1/3) = c^(1/3). Moreover, there is an infinite number of solutions to FLTR for general n. E.g. 8^(1/3) + 27^(1/3) = 125^(1/3). However, in conjunction with a^2+b = c, I could not find any nontrivial solutions. Perhaps there is another formula that will yield solutions.

Define FLTR as Fermat's Last Theorem with rational exponents. Consider x + y = x + y. Then (x^n)^(1/n) + (y^n)^(1/n) = ((x+y)^n)^(1/n) or FLTR For n=2 we have, (x^2)^(1/2) + (y^2)^(1/2) = ((x+y)^2)^(1/2) So a = x^2, b = y^2 and c = (x+y)^2 Then a^2 + b = c => x^4 + y^2 = (x+y)^2 x^4 + y^2 = x^2 + 2xy + y^2 y = (x^3-x)/2 (x+y)^2 is the generating function for this sequence for x=0,1,2,3,..

For a=9,b=144,c=225, 9^(1/2) + 144^(1/2) = 225^(1/2) and 9^2 + 144 = 225. So

c = 225 is the 4th entry in the sequence.

(PARI) flt2(n, p) = { local(a, b); for(a=0, n, b = (a^3-a)/2; print1(b^2", ") ) }

Sequence in context: A030484 A038692 A017330 this_sequence A065779 A095241 A061839

Adjacent sequences: A135506 A135507 A135508 this_sequence A135510 A135511 A135512

nonn

Cino Hilliard (hillcino368(AT)hotmail.com), Feb 09 2008