TL;DR

On with TASK #2 from The Weekly Challenge #148. Enjoy!

The challenge

Write a script to generate first 5 Cardano Triplets.

A triplet of positive integers (a,b,c) is called a Cardano Triplet if it satisfies the below condition.

\[\sqrt[3]{a + b \sqrt{c}} + \sqrt[3]{a - b \sqrt{c}} = 1\]

Example

(2,1,5) is the first Cardano Triplets.

The questions

Is the internet a valid source? Because I’m definitely using it!

To be more serious, I don’t know what does it mean to generate the first Cardano Triplets. What’s the ordering supposed to be? Lowest sum of the three values $a$, $b$, and $c$? Something else?

The solution

It seems that our fine host discovered Project Euler and gives us (well… me) the occasion to cheat a bit and look for solutions around.

It’s clear that doing square and cube roots is going to spoil all the fun with integers in a computer, so the solution MUST be something that can be solved through integer-only maths. And there it is, the characterization I was looking for.

Alas, this week is finding me particularly lazy, so I’ll take the extremely simple route and adopt this:

with $b > 0$ integer. This means that, from the characterization, I’m assuming that $b = k + 1$, even though it might be that $8 k + 5$ is a square too… in which case I might have another candidate triple. Whatever.

At this point, we just have to iterate over the needed values for $b$. Let’s get Perlish first:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

my $n = shift // 5;
for my $b (1 .. $n) {
   my $a = 3 * $b - 1;
   my $c = 8 * $b - 3;

   my $sqrt = $b * sqrt($c);
   my $first = ($a + $sqrt) ** (1/3);
   my $second = ($sqrt - $a) ** (1/3);
   my $result = $first - $second;

   say "($a, $b, $c) -> $result";
}

The whole calculation for the $result seems to be kind with us:

$ perl perl/ch-2.pl 
(2, 1, 5) -> 1
(5, 2, 13) -> 1
(8, 3, 21) -> 1
(11, 4, 29) -> 1
(14, 5, 37) -> 1

Let’s move on to Raku now, with pretty much the same implementation, apart the different syntax:

#!/usr/bin/env raku
use v6;
sub MAIN (Int:D $n = 5) {
   for 1 .. $n -> $b {
      my $a = 3 * $b - 1;
      my $c = 8 * $b - 3;

      my $sqrt = $b * $c.sqrt;
      my $first = ($a + $sqrt) ** (1 / 3);
      my $second = ($sqrt - $a) ** (1 / 3);
      my $result = $first - $second;

      "($a, $b, $c) -> $result".put;
   }
}

This time, anyway, the control calculations do not help us understand:

$ raku raku/ch-2.raku 
(2, 1, 5) -> 0.9999999999999999
(5, 2, 13) -> 1.0000000000000002
(8, 3, 21) -> 1
(11, 4, 29) -> 1
(14, 5, 37) -> 1

Well, at least we understood it correctly in the beginning… a solution involving square and cube roots would not bring us too far.

Stay safe folks!