There is, indeed, a flaw in Constant's proof from http://www.coolissues.com/mathematics/Beal/beal.htm. In his "Disproof of Beal's Conjecture," after he applies the binomial theorem, he gets three results. His error comes in the second one.
He is right in that (1+y^b/x^a)^(1/m) converges to an irrational number, but he errs in saying that "z in equation (4) is an irrational number for any x^(a/m)," because there exist some x^(a/m), such that multiplying the two numbers will yield a rational number. The assumption is that multiplying an irrational number by any number will stay irrational, but there are pairs of irrational numbers that, when multiplyed, are rational (e.g. pi * 1/pi = 1).
Take the simple example of Beal's conjecture, 3^3+6^3=3^5. Inputting these variables into his binomial expansion form, we get 3=6^(3/5)*(1+3^3/6^3)^(1/5). The parentheses term,(1+27/216)^(1/5), is irrational, but multiplying by 6^(3/5), another irrational number, results in a solution of 3.
It's a simple flaw, but it means that the solution isn't as close as it may seem. |
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