SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

Answer: $0$.

Let $x=ac+3bd$, $y=ad-bc$. We have $$\begin{array}{rl}37=5^2+3\cdot 2^2& =(ace+3adf-3bcf+3bde{)}^2+3(acf-ade+bce+3bdf{)}^2\\ & =(ex+3fy{)}^2+3(fx-ey{)}^2=(e^2+3f^2)(x^2+3y^2)\\ & =(e^2+3f^2)((ac+3bd{)}^2+3(ad-bc{)}^2)\\ & =(e^2+3f^2)(c^2+3d^2)(a^2+3b^2).\end{array}$$ 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: $a=1$, $b=0$, $c=1$, $d=0$, $e=5$, $f=2$.