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 Iâ€™m definitely using it!

To be more serious, I donâ€™t know what does it mean to generate the first Cardano Triplets. Whatâ€™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 (wellâ€¦ me) the occasion to cheat a bit and look for solutions around.

Itâ€™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 Iâ€™ll take the extremely simple route and adopt this:

$a = 3 b - 1 \\ c = 8 b - 3$

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

At this point, we just have to iterate over the needed values for $b$. Letâ€™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:

$perl perl/ch-2.pl (2, 1, 5) -> 1 (5, 2, 13) -> 1 (8, 3, 21) -> 1 (11, 4, 29) -> 1 (14, 5, 37) -> 1  Letâ€™s move on to Raku now, with pretty much the same implementation, apart the different syntax: #!/usr/bin/env raku use v6; sub MAIN (Int:D$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 beginningâ€¦ a solution involving square and cube roots would not bring us too far.

Stay safe folks!

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