On 06/02/2018 08:58, Christian Boyer wrote:
From www.multimagie.com/English/SquaresOfCubes.htm
About your conjecture 1: "Euler and Legendre demonstrated that x^3 + y^3 = k * z^3 is impossible with distinct integers, for k = 1, 2, 3, 4, 5."
About your conjecture 2: "Legendre showed also that x^4 + y^4 = 2 * z^2 is impossible if x != y."
Keith's conjecture 3 was: "No two positive integers have both a geometric mean that's an integer and a quadratic mean that's an integer." That is, we are asking whether we can have x^2+y^2 = 2z^2 and also xy = t^2 with x,y,z,t positive integers. (And with x,y distinct.) The latter means that x=a^2.r and y=b^2.r for some positive integers a,b,r. Then the former says (a^2r)^2 + (b^2r)^2 = 2z^2 or a^4 + b^4 = 2(z/r)^2, and clearly z/r must be an integer in this case; call it c. So a^4 + b^4 = 2c^2. If this has solutions with a,b distinct then consider a minimal one. Any common factor of a,b would lead to a smaller example (dividing a,b,c by d,d,d^2) so a,b are coprime. Now according to https://math.stackexchange.com/questions/339410/x4y4-2z2-has-only-solution-x... (warning: I haven't checked the reasoning, though it seems to be doing the right sort of thing) the only solution has a=b=1 and we're done: Keith's conjecture 3 is correct. -- g