# ETOOBUSY ðŸš€ minimal blogging for the impatient

# PWC148 - Cardano Triplets

**TL;DR**

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

# The challenge

Write a script to generate first 5 Cardano Triplets.

\[\sqrt[3]{a + b \sqrt{c}} + \sqrt[3]{a - b \sqrt{c}} = 1\]A triplet of positive integers (a,b,c) is called a Cardano Triplet if it satisfies the below condition.

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:

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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