Let $c_1<c_2<c_3$ be natural numbers and $$C_n=\prod_{i=1}^3(nc_i+1)-1\;\;\;\;\;(n\in\mathbb{N})$$ I want to show that it holds $$\forall n\forall i : nc_i\mid C_n\;\;\;\Leftrightarrow\;\;\;\exists c\in6\mathbb{N}:c_i=ic$$ The direction $(\Leftarrow)$ is easy, but I don't know how one can see $(\Rightarrow)$


Let $(x,y,z)$ be a cyclic permutation of $(1,2,3)$. You want that for all $n$

$$nc_x \mid \prod_{i=1}^3 (nc_i + 1) - 1.$$

Expanding the product, you get

$$nc_x \mid nc_x[(nc_y+1)(nc_z+1)] + [(nc_y+1)(nc_z+1)-1].$$

Evidently $nc_x$ divides the first summand, so we only need be concerned with the second:

$$(nc_y+1)(nc_z+1)-1 = n^2 c_yc_z + nc_y + nc_z.$$

We can divide by $n$, and obtain the requirement

$$\bigl(\forall n\bigr)\bigl(c_x \mid [nc_yc_z + c_y + c_z]\bigr).$$

Looking at the right hand term for two consecutive $n$, we deduce $c_x \mid c_y c_z$, and subtracting $nc_yc_z$ (which we now know is divisible by $c_x$), we also find $c_x \mid c_y + c_z$.

Thus we arrive at the conditions

$$c_x \mid c_yc_z \land c_x \mid c_y + c_z,$$

for all cyclic permutations $(x,y,z)$ of $(1,2,3)$.

From $c_3 \mid c_1+c_2$ and $c_1 < c_2 < c_3$ we obtain $c_1 + c_2 < 2c_3$, so we must have $c_3 = c_1+c_2$. Then we find

$$c_2 \mid c_1 + c_3 = 2c_1 + c_2 \implies c_2 \mid 2c_1. $$

From that we deduce $c_2 = 2c_1$, so indeed $c_i = i\cdot c_1$ for $i = 1,2,3$.

It remains to see that $6\mid c_1$.

$$\begin{gather} c_2 \mid c_1c_3 \iff 2c_1 \mid 3 c_1^2 \iff 2\mid 3c_1 \iff 2\mid c_1,\\ c_3 \mid c_1c_2 \iff 3c_1 \mid 2 c_1^2 \iff 3\mid 2c_1 \iff 3\mid c_1 \end{gather}$$

yield that.


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