Integers with 3 Prime Factors or More

Proof: Let $F$ be the set of prime factors of $n$. Since $n=pqrc$, $p$, $q$, and $r$ are three distinct prime numbers, we have $|F|\ge 3$.

Assume, for contradiction, that for every proper factor $m$ of $n$, $1<m<n$, we have $m\ |\ x$ or $m\ |\ y$.

For each prime number $f\in F$, consider $s_f=\dfrac{n}{f}$. Since $f\ge 2$, we have $1<s_f<n$. Therefore, $s_f$ is a proper factor of $n$. Hence, for each $f\in F$, either $s_f\ |\ x$ or $s_f\ |\ y$.

Let $$F_x={\ f\in F : s_f\ |\ x\ },\ \ \ \ \ \ \ F_y={\ f\in F : s_f\ |\ y\ }$$ We have $F=F_x\cup F_y$.

If $|F_x|\ge 2$, then there are two distinct prime factors, $f$ and $f’$ in $F_x$,
such that $s_f\ |\ x$ and $s_{f’}\ |\ x$. Since $f$ and $f’$ are two distinct prime factors of $n$, then $n=f\cdot f’\cdot M$, where $M$ is a positive integer. Then
$$s_f=\dfrac{n}{f}=\dfrac{f\cdot f’\cdot M}{f}=f’\cdot M \ \ \ \ \ \ \
s_{f’}=\dfrac{n}{f’}=\dfrac{f\cdot f’\cdot M}{f’}=f\cdot M$$
Because $f$ and $f’$ are distinct prime factors of $n$, we have
$$lcm(s_f, s_{f’})=lcm(f’\cdot M, f\cdot M)=f\cdot f’\cdot M=n$$

Because $s_f\ |\ x$ and $s_{f’}\ |\ x$, we have $lcm(s_f,s_{f’})\ |\ x$, i.e $n\ |\ x$. Therefore $x\ge n$, contradicting $x\le n-1$. Therefore $|F_x|\le 1$. Similarly, $|F_y|\le 1$.

Thus, $|F|=|F_x\cup F_y|\le |F_x|+|F_y|\le 1+1=2$, contradicting $|F|\ge 3$.

Therefore, the assumption is false, the claim is proved.