Find all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $a<b<c$

Question

Find all integers $x$ that can't be written as $x=a+b+c$, where $a$ divides $b$ and $b$ divides $c$ with $a<b<c$

This implies that $a$ divides $c$.

I can also rewrite $c=zb$ and $b=ya$ so $c=zya$ with $z\neq0$ and $y\neq0$

So $x=a(1+z+zy)$

$z$ cannot be equal to $1$ because $c=zb$ and $b<c$

$y$ cannot be equal to $1$ because $b=ya$ and $a<b$

So I can state that none of the prime number can be write under that form if $a\neq1$.

But what then?

Answer 1

There is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.

You should probably consider the case when $a=1$ first. There is a solution to $x-1=z(1+y)$ when $x-1$ is non-prime and bigger than $4$ (since $1+y>2$.)

On to the general equation.

If $X$ has an odd factor $d>5$, then $d-1=2\left(1+\frac{d-3}{2}\right)$, and $ X=\frac{X}{d} + \frac{2X}{d} + \frac{2X(d-3)}{d}$.

So we can reduce the question to $X$ which can be written as $2^n$, $3\cdot 2^n$ and $5\cdot 2^n$.

There is no solution for $1,2,3,4,5,6$.

Since $2\cdot 5-1=3\cdot 3$, we get that $5\cdot 2^n = 2^{n-1}+3\cdot 2^{n-1}+3\cdot 2^{n-1}$.

Since $3\cdot 2^5-1=5\cdot 19$, we have that:

$$3\cdot 2^{n+5}=2^{n}+5\cdot 2^n + 18\cdot 5\cdot 2^n$$

$$x=2^{n+4}=2^n + 5\cot 2^{n+1} + 5\cdot 2^{n+2}$$

Left to check: $x=8,12,24,48$. $48=3(2^4)=3+15+30$.

There is no solution for $8$, since if $c>4$ then $c\geq 6$ and $a+b+c>8$.

For $x=12$, we can't have $a=1$ since $x-1$ is prime. If $a>1$ then $12/a \leq 6$ and $1+y+yz=12/a$ has no solution.

$x=24$ is the last case. For all divisors $a\mid x$, you have either $x/a-1$ is prime, or $x/a-1<6$.

So there is no solution for $x=1,2,3,4,5,6,8,12,24$ and solutions for all other $x$.

Answer 2

Instead of writing in terms of $a,b$ and $c$, we know $b = ka$ and $c = mb = kma$. So we are looking for all numbers of the form $(1 +k + mk)*a = (1+k(1+m))*a = (1 + k*n)*a$ where $a,k,n$ are all positive integers and $k > 1;n > 2$.

In other words $x = (1 + v)*a$ where $v$ is a composite; $v \ge 6$.

So $x$ can be an composite number where one of its factors or itself is one more than a composite number at least as large as 6.

These numbers are very boring. It can include primes. Indeed it includes all primes except 2,3 and 5.

In fact, the only numbers it can't include are numbers $x$ is 1 more than a prime, such that $x$ only factors to terms that are both one more than primes. Which is impossible as those factors would be composite so we can find other factors. I'm guessing these types of numbers are finite and I wouldn't be surprised if the largest was 12. But I'm speaking off the top of my head.

Let's see... $8 = 1 + 7$ is not in this form. $12 = 1 + 11 = 2*6$ is not in this form. $14 = 2*7 = 2(1 + 6)$ is in this form. $24 = 1 + 23=2(1 + 11)= 3(1 + 7)=4*6$ is not of this form. $48 = 1 + 47 = 2*(1 + 23)= 3*(1 + 15)$ is off this form. To go higher we need $m =1 + p= 2(1+ p); 1+p = 2(1 + q);$ etc. which I don't think is possible for larger primes. I'm pretty sure $24$ is the largest number not in this form.

Answer 3

I can't see any restrictions against negative integers. So $a$ could be negative. And $b$ and $c$ must be positive for the requirement $a<b<c$. Then we can achieve any integer $n>1$ using $a=-1, b=1, c=n$. We can't achieve any integer less than or equal to $1$ (since $b$ must be positive with $b+a\ge 0$.)