Please check if the following proof that for every prime $a$, the number of carmichael numbers $N=abc$ , where $b$ and $c$ are also prime, is finite, is correct!
First of all, we have
$$a-1|abc-1$$ $$b-1|abc-1$$ $$c-1|abc-1$$
which implies
$$a-1|bc-1$$ $$b-1|ac-1$$ $$c-1|ab-1$$
Considering the last two conditions, we get the equation system
$$ac-1=mb-m$$ $$ab-1=nc-n$$
or
$$mb-ac=m-1$$ $$-ab+nc=n-1$$
Neither $m$ nor $n$ can be disivible by $a$, so $mn-a^2\ne 0$ must hold.
The solution of the system is
$$b=\frac{mn+(a-1)n-a}{mn-a^2}$$ $$c=\frac{mn+(a-1)m-a}{mn-a^2}$$
In particular, we get
$$(a-1)n-a+a^2=(a-1)(a+n)\ge mn-a^2$$ $$(a-1)m-a+a^2=(a-1)(a+m)\ge mn-a^2$$
and finally
$$2a^2-a\ge n(m-a+1)$$ $$2a^2-a\ge m(n-a+1)$$
implying $1\le m,n\le 2a^2-1$
Is this proof correct ?
Is there a better way to determine all pairs (b,c) ?
If more than $3$ prime factors are allowed, is the number of carmichael numbers containing a fixed prime factor $a$, finite ?