rafaello wrote: Distinct positive integers $a, b, c, d$ satisfy $$\begin{cases} a \mid b^2 + c^2 + d^2,\\ b\mid a^2 + c^2 + d^2,\\ c \mid a^2 + b^2 + d^2,\\ d \mid a^2 + b^2 + c^2,\end{cases}$$and none of them is larger than the product of the three others. What is the largest possible number of primes among them? 3 primes are possible ($a=2,b=3,c=13,d=26$). Suppose $a>b>c>d$ are different primes. We have \begin{align*} a\mid a^2+b^2+c^2+d^2\\ b\mid a^2+b^2+c^2+d^2\\ c\mid a^2+b^2+c^2+d^2\\ d\mid a^2+b^2+c^2+d^2\\ abcd\mid a^2+b^2+c^2+d^2\\ kabcd=a^2+b^2+c^2+d^2 \end{align*}for some positive integer $k$. By https://artofproblemsolving.com/community/q1h2713763p23603728 we have $k\leq4$. Modulo 4 gives $k=4$ since $a,b,c,d$ are odd. Thus \[ a^2+b^2+c^2+d^2=4abcd=abcd+3abcd>a^2+45ab>a^2+b^2+c^2+d^2 \]which is impossible.

@above unfortunately, 62 is larger than 2.3.7 (I ran into the same issue)

A much shorter solution: example four three above, assume there is a quadruple with four can be found; and let $a<b<c<d$ wlog. Then, $abcd\mid a^2+b^2+c^2+d^2$. Clearly if $a=2$ then $a^2+b^2+c^2+d^2$ is odd, whereas $abcd$ is even, absurd. Hence, $a,b,c,d$ are all odd. Setting $a^2+b^2+c^2+d^2 = \ell \cdot abcd$, we find $a^2+b^2+c^2+d^2\equiv 0\pmod{4}$, and since $abcd$ is odd, we get $4\mid \ell$. Thus $\ell\ge 4$. Notice, however, that \[ \ell = \sum_{{\rm sym}} \frac{a}{bcd}<4, \]as $d\le abc$ and for every $x\in\{a,b,c\}$, $x<P/x$ for $P=abcd$.