[Algebra]
Problem: Let $i=\sqrt{-1}$. The complex number $z=-142+333 \sqrt{5} i$ can be expressed as a product of two complex numbers in multiple different ways, two of which are $(57-8 \sqrt{5} i)(-6+5 \sqrt{5} i)$ and $(24+\sqrt{5} i)(-3+14 \sqrt{5} i)$. Given that $z=-142+333 \sqrt{5} i$ can be written as $(a+b \sqrt{5} i)(c+d \sqrt{5} i)$, where $a, b, c$, and $d$ are positive integers, compute the lesser of $a+b$ and $c+d$.

Solution: Multiply each of the given parenthesized expressions by its complex conjugate to obtain

$$
\begin{aligned}
142^{2}+5 \cdot 333^{2} & =\left(57^{2}+5 \cdot 8^{2}\right)\left(6^{2}+5 \cdot 5^{2}\right) \\
& =\left(24^{2}+5 \cdot 1^{2}\right)\left(3^{2}+5 \cdot 14^{2}\right) \\
& =\left(a^{2}+5 b^{2}\right)\left(c^{2}+5 d^{2}\right) .
\end{aligned}
$$

The expression on the second line is equal to $581 \cdot 989=7 \cdot 83 \cdot 23 \cdot 43$ (one can perhaps factor 989 a little faster by noting that 23 divides $6^{2}+5 \cdot 5^{2}=7 \cdot 23$ but not 581 , so it must divide 989 ). Thus $a^{2}+5 b^{2}$ and $c^{2}+5 d^{2}$ must be a factor pair of this number. It is not possible to express $1,7,23,43$, or 83 in the form $x^{2}+5 y^{2}$ for integers $x, y$.

Let $N=a^{2}+5 b^{2}$, and without loss of generality, assume that 7 divides $N$. From the above analysis, $N$ must be $7 \cdot 23,7 \cdot 43$, or $7 \cdot 83$. By direct computation of checking all positive integers $b$ less than $\sqrt{\frac{N}{5}}$, the only possibilities for $(a, b)$ are:

- when $N=7 \cdot 23$, either $(9,4)$ or $(6,5)$;
- when $N=7 \cdot 43$, either $(16,3)$ or $(11,6)$; and
- when $N=7 \cdot 83$, either $(24,1)$ or $(9,10)$.

Next, observe that

$$
\frac{-142+333 \sqrt{5} i}{a+b \sqrt{5} i}=\frac{(-142 a+1665 b)+(333 a+142 b) \sqrt{5} i}{N}
$$

must equal $c+d \sqrt{5} i$, so $N$ must divide $-142 a+1665 b$ and $333 a+142 b$. But

- 7 does not divide $333 \cdot 9+142 \cdot 4$ or $333 \cdot 6+142 \cdot 5$;
- 43 does not divide $333 \cdot 16+142 \cdot 3$; and
- 83 does not divide $333 \cdot 9+142 \cdot 10$.

Thus the only candidates are $(a, b)=(11,6)$ and $(a, b)=(24,1)$. Note that $(24,1)$ yields the second factorization given in the problem statement, which has a negative real part in one of its factors. Thus the only remaining candidate for $(a, b)$ is $(11,6)$, which yields $(c, d)=(28,15)$, thus the answer is $11+6=\mathbf{1 7}$.

Final answer: 17