If a positive integer $n$ is equal to $a + b + c$ where $a | b$, $b | c$ and $a \lt b \lt c$, let it be called "faithful". Prove that nearly all numbers are faithful and list the non-faithful numbers.
Here's where I am:
Let the number be $n$, $b = ap$ and $c = bq = apq$ where $p ,q \in Z_+$. Let the set of faithful numbers be $F$.
$$n = a + b + c$$ $$\therefore n = a + ap + apq$$
Since $1 \lt p \lt q$ and $p, q \in Z_+$, $n \ge 7$. Hence $1, 2 \ldots 7 \notin F$.
Now what?
On
Lemma $n \geq 7$ can be written in the form $1+p+pq$ where $1< p < q$ if and only if $n-1 \neq p, p^2+p$ for $p$ prime.
The proof follows immediately from:
$$n=1+p+pq \Leftrightarrow n-1=p(q+1)$$
This proves that all numbers $n \geq 7$ which are not of the forms $p+1$ or $p^2+p+1$ are faitful.
Lemma 2: If $n=p^2+p+1$ then $n$ is faithful.
Proof: $a=1, b=2, c=p^2+p-2$.
This shows that $F \subset \{ 1,2,3,4,5,6,7 \} \cup \{ p+1| p \, \mbox{prime} \}$.
To complete the proof, all you need to do is to observe that
Lemma 3: If $n=p+1 > 7$ and $n$ is faitfull then $\frac{n}{2}$ is also faitful.
Proof assume by contradiction that $\frac{n}{2}=a+b+c$ then $n=2a+2b+2c$ contradiction.
Now, all you need is brute force it. The following lemma tells you how:
Lemma 4 If for some $n$ there is no faithful number between $n$ and $2n$, then there is no faithful number greater than $n$.
Proof By contradiction. Assume there is some $m>n$ faithful. Then $m=p+1>2n$ and hence by Lemma 3 we have $\frac{m}{2}$ is also faithful. Repeat until the first time you get a number under $2n$, that number must be between $n$ and $2n$ contradiction.
The brute force Look for all the faithfull numbers between $8$ and $16$ of the type $p+1$. If none, done by lemma 4. If you find any, pick the largest of them $k$ and look for all the faithfull numbers between $(k+1)$ and $2(k+1)$ of the type $p+1$. Repeat until you find none.
This works only if you are sure there are only finitely many faithful numbers.
Let $a=2^k$ for some $k\ge 0$, and let $p=2$. Then $a+ap+apq=2^k(3+2q)$, where we must have $q\ge 2$. $\{3+2q:q\ge 2\}$ is the set of odd numbers greater than $5$, so every number of the form $2^k(2m+1)$ with $m\ge 3$ is faithful. The potential exceptions are therefore the numbers of the form $2^k,2^k\cdot3$, and $2^k\cdot 5$ for $k\ge 0$.
Suppose that one of these numbers can be written in the form $a+ap+apq$ for some $a\ge 1$ and $p,q\ge 2$. Then it can be written as $a(1+p+pq)$, where $1+p+pq\ge 7$.
If $k\ge 4$, $2^k=2^{k-4}\cdot16=2^{k-4}(1+3+3\cdot4)$, so we may take $a=2^{k-4}$, $p=3$, and $q=4$. Thus, $2^k$ is faithful for $k\ge 4$, and we need consider only $2,4$, and $8$ individually; all are easily seen to be unfaithful.
Similarly, $2^k\cdot3=3\cdot2^{k-4}(1+3+3\cdot4)$, so we may take $a=3\cdot2^{k-4}$, $p=3$, and $q=4$ and catch all numbers of the form $2^k\cdot3$ except $12$ and $24$, and $2^k\cdot5=5\cdot2^{k-4}(1+3+3\cdot4)$, so we may take $a=5\cdot2^{k-4}$, $p=3$, and $q=4$ and catch all numbers of the form $2^k\cdot 5$ except $10,20$, and $40$.
We’ve now shown that all positive integers are faithful except $1,2,3,4,5,8$, and possibly $10,12,20,24$, and $40$. $10=1+3+6$ is faithful. It’s not hard to check that $12$ is not faithful: its only factor larger than $6$ is itself, so $a$ would have to be $1$, and $11=p(1+q)$ would have to be composite. $20=2\cdot20=2+6+12$ is faithful, and similarly $40$ is faithful. That leaves only $24$ to be checked, and I’ll leave it to you.