Project Euler Problem 251
A triplet of positive integers (a, b, c) is called a Cardano Triplet if it satisfies the condition: For example, (2,1,5)
Solution
Answer: 18946051
Let
$$x=\sqrt[3]{a+b\sqrt c}, \qquad y=\sqrt[3]{a-b\sqrt c}.$$
We are given
$$x+y=1.$$
Also,
$$x^3+y^3=(a+b\sqrt c)+(a-b\sqrt c)=2a.$$
Using
$$x^3+y^3=(x+y)^3-3xy(x+y),$$
and $x+y=1$,
$$2a = 1-3xy.$$
So
$$xy=\frac{1-2a}{3}.$$
Since
$$xy=\sqrt[3]{a^2-b^2c},$$
we cube both sides:
$$a^2-b^2c=\left(\frac{1-2a}{3}\right)^3.$$
For the right-hand side to be integral, $2a-1$ must be divisible by $3$. Write
$$a=3k+2 \qquad (k\ge0).$$
Then
$$\frac{1-2a}{3}=-(2k+1),$$
hence
$$b^2c = a^2+(2k+1)^3.$$
Substituting $a=3k+2$,
$$b^2c = (3k+2)^2+(2k+1)^3.$$
A crucial factorization appears:
$$(3k+2)^2+(2k+1)^3 = (k+1)^2(8k+5).$$
Thus every Cardano triplet is parameterized by
$$a=3k+2,$$
and
$$b^2c=(k+1)^2(8k+5).$$
For each $k$, every valid $b$ is such that $b^2$ divides $(k+1)^2(8k+5)$, with
$$c=\frac{(k+1)^2(8k+5)}{b^2},$$
subject to
$$a+b+c\le 110{,}000{,}000.$$
A carefully optimized divisor enumeration reproduces the check value $149$ for $a+b+c\le1000$, matching the problem statement, and then evaluates the full limit.
Answer: 18946051