Just to confirm: Are you really sure about the definition of $\omega(n)$? I am asking because this is not how it is commonly used. Usually, $\omega(n)$ denotes the number of distinct prime factors, while $\Omega(n)$ counts the prime factors with multiplicity, i.e. $\omega(12)=2$, but $\Omega(12)=3$.

Tintarn wrote: Just to confirm: Are you really sure about the definition of $\omega(n)$? I am asking because this is not how it is commonly used. Usually, $\omega(n)$ denotes the number of distinct prime factors, while $\Omega(n)$ counts the prime factors with multiplicity, i.e. $\omega(12)=2$, but $\Omega(12)=3$. Thanks. In China, there isn’t a clear definition of this function, sorry

Answer: $3$ To see that $3$ is achievable take $(a,b,c,d)=83\cdot (6,5,4,3)$. Then $$ab+bc+ca+d^2=83^2\cdot(6\cdot 5+5\cdot 4+4\cdot 6+3^2)=83^3|(a+b)(b+c)(c+a)$$Clearly by size $ab+bc+ca+d^2$ can not be prime. Now assume that $ab+bc+ca+d^2$ is divisible by $2$ primes $p$ and $q$. Then forget about $c<b<a$ and WLOG assume $$ab+bc+ca+d^2|(a+b)(c+a)=ab+bc+ca+a^2$$Clearly $a+d<pq$. Also we must have $a^2 \equiv d^2 \pmod{p}$ and $a^2\equiv d^2 \pmod{q}$. If $a\equiv \pm d \pmod{pq}$ then we have a size contradiction so WLOG assume that $a\equiv- d\pmod{p}$ and $a\equiv d \pmod{q}$. Also WLOG assume that $p|(a+b)$. Then $a\equiv -b \pmod{p}$ so $b\equiv d \pmod{p}$ so we must have $p<b$. By size we must have that $q|(a+c)$. Then $$q=\frac{ab+bc+ca+d^2}{p}>\frac{ab+bc+ca+d^2}{b}>a+c$$A contradiction, as desired.

Very elegant official solution: Note that $(a+b+c+d)(ab+bc+ca+d^2)=(a+b)(b+c)(c+a)+(a+d)(b+d)(c+d)$. The result follows by size. All four members of our team solved this one, but none using this short and elegant solution, but rather similar to the one above.

Another nice solution $ab+bc+ca+d^2|(a+b)(b+c)(c+a)=(ab+ac+bc+b^2)(a+c)$ $ab+bc+ca+d^2|(b^2-d^2)(c+a)=(b-d)(b+d)(c+a)=(b-d)(bc+ba+cd+da)$ Then $ab+bc+ca+d^2|-(b-d)(-ac-d^2+cd+da)=(b-d)(c-d)(a-d)$ So from $ab+bc+ca+d^2|(a‐d)(b-d)(c-d)$ due to size we notice that LHS cannot be coprime with any of $(a-d)$ , $(b-d)$ or $(c-d)$ And that gives $\omega(n)>2$ And the construction same as above So we get $\omega(n)=3$

Here is my solution $\omega(n)=3$ Where Rasid from alawais?