[math-fun] Power-mean conjectures
I've been looking at pairs of positive integers which have integer power means, including the harmonic, geometric, arithmetic, quadratic (RMS), cubic, and quartic means. Conjecture 1: No two positive integers have a cubic mean that's an integer. Conjecture 2: No two positive integers have a quartic mean that's an integer. Conjecture 3: No two positive integers have both a geometric mean that's an integer and a quadratic mean that's an integer. All three are true for all possible pairs of positive integers less than ten thousand. Are they true for all larger integers too? Other than that, all possible combinations appear to exist. For instance 5 and 45 have a harmonic mean of 9, a geometric mean of 15, and an arithmetic mean of 25. 4 and 28 have a harmonic mean of 7, an arithmetic mean of 16, and a quadratic mean of 20. And plenty of pairs have just 0, 1, or 2 means that are integers, in all possible combinations that don't violate my conjectures. Does anyone have a proof or counterexample for any of my conjectures? Thanks.
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." Christian. -----Message d'origine----- De : math-fun [mailto:math-fun-bounces@mailman.xmission.com] De la part de Keith F. Lynch Envoyé : mardi 6 février 2018 06:16 À : math-fun@mailman.xmission.com Objet : [math-fun] Power-mean conjectures I've been looking at pairs of positive integers which have integer power means, including the harmonic, geometric, arithmetic, quadratic (RMS), cubic, and quartic means. Conjecture 1: No two positive integers have a cubic mean that's an integer. Conjecture 2: No two positive integers have a quartic mean that's an integer. Conjecture 3: No two positive integers have both a geometric mean that's an integer and a quadratic mean that's an integer. All three are true for all possible pairs of positive integers less than ten thousand. Are they true for all larger integers too? Other than that, all possible combinations appear to exist. For instance 5 and 45 have a harmonic mean of 9, a geometric mean of 15, and an arithmetic mean of 25. 4 and 28 have a harmonic mean of 7, an arithmetic mean of 16, and a quadratic mean of 20. And plenty of pairs have just 0, 1, or 2 means that are integers, in all possible combinations that don't violate my conjectures. Does anyone have a proof or counterexample for any of my conjectures? Thanks. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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
participants (3)
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Christian Boyer -
Gareth McCaughan -
Keith F. Lynch