HP50g simplifing a root

[Albert Chan]

(10-05-2020 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!
The checking is absolut necessary.

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.

(09-30-2020 02:22 AM)Albert Chan Wrote:  For general case, to solve for all a, b:

      ³√(n + m√k) = a + b√k

(a + b√k)³ = a³ + 3a²b√k + 3ab²k + b³k√k

n = a³ + 3ab²k       → n/a - a² = 3b²k
m = 3a²b + b³k      → 3m/b - 9a² = 3b²k

Since a|n, 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) ⇒ (3|c) ⇒ (b = c/3 = integer)

Assumption were wrong, b must be integer.      QED