Distinct positive integers $a, b, c, d$ satisfy

$$
\left\{\begin{array}{l}
a \mid b^{2}+c^{2}+d^{2}, \\
b \mid a^{2}+c^{2}+d^{2}, \\
c \mid a^{2}+b^{2}+d^{2}, \\
d \mid a^{2}+b^{2}+c^{2},
\end{array}\right.
$$

and none of them is larger than the product of the three others. What is the largest possible number of primes among them?
At first we note that the given condition is equivalent to $a, b, c, d \mid a^{2}+b^{2}+c^{2}+d^{2}$.

It is possible that three of the given numbers are primes, for example for $a=2, b=3, c=13$ and $d=26$. In this case $2^{2}+3^{2}+13^{2}+26^{2}=13 \cdot 66$ which is divisible by all four given numbers. Furthermore we will show that it is impossible that all four of them are primes.

Let us assume that $a, b, c$ and $d$ are primes. As the sum $a^{2}+b^{2}+c^{2}+d^{2}$ is divisible by each of them then it is divisible also by their product $a b c d$. If one of the primes is equal to 2, then we obtain a contradiction: the sum of four squares is odd, but its divisor $a b c d$ is even. Therefore all four primes are odd, and $a^{2}+b^{2}+c^{2}+d^{2}=0(\bmod 4)$. Hence $a^{2}+b^{2}+c^{2}+d^{2}$ is divisible by $4 a b c d$ which leads to a contradiction as it is easy to see that $a^{2}+b^{2}+c^{2}+d^{2}<4 a b c d$. Indeed, this is equivalent to

$$
\frac{a}{b c d}+\frac{b}{a c d}+\frac{c}{a b d}+\frac{d}{a b c}<4
$$

which is true as none of the numbers exceed the product of the other three and equality can hold only for the largest of the four.