HP50g simplifing a root

10052020, 05:01 PM
Post: #13




RE: HP50g simplifing a root
(10052020 11:36 AM)peacecalc Wrote: thank you for critics. Of course I forgot to implement the testing procedure being shure that I found a correct solution! This checking is the reason my previous posts had different arguments requirement. find_ar(n, m) // note, no k's It returned simplified a + b√k, but the user *must* check √k matches. find_cbrt(n, m, k) // we required k to build the cubic It solved the cubic, checked a is integer, but assumed b is also integer. Here is a proof that shows b is indeed an integer. (09302020 02:22 AM)Albert Chan Wrote: For general case, to solve for all a, b: Since an, n/a  a² = 3b²k = integer Assume b ≠ integer, but b = c/3, where c = integer. m = 3a²b + b³k = a²c + c³k/27 = integer Assuming √k is fully reduced, k can not have factor of 27 = 3³ (c³k/27 = integer) ⇒ (3c) ⇒ (b = c/3 = integer) Assumption were wrong, b must be integer. QED 

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