Problem 1666

Answer: $0$.

Let $x = ac + 3bd$, $y = ad - bc$. We have
$$\begin{align*} 37 = 5^2 + 3 \cdot 2^2 &= (ace + 3adf - 3bcf + 3bde)^2 + 3(acf - ade + bce + 3bdf)^2 \\ &= (ex + 3fy)^2 + 3(fx - ey)^2 = (e^2 + 3f^2)(x^2 + 3y^2) \\ &= (e^2 + 3f^2)((ac + 3bd)^2 + 3(ad - bc)^2) \\ &= (e^2 + 3f^2)(c^2 + 3d^2)(a^2 + 3b^2). \end{align*}$$
Since the number $37$ is prime, we obtain $\{a^2 + 3b^2, c^2 + 3d^2, e^2 + 3f^2\} = \{1, 1, 37\}$. However, if the integers $m$ and $n$ satisfy the equation $m^2 + 3n^2 = 1$, then $n = 0$. Therefore, two of the numbers $b, d, f$ are equal to zero and, therefore, $abcde = 0$. Here is one of the solutions to the system: $a = 1$, $b = 0$, $c = 1$, $d = 0$, $e = 5$, $f = 2$.