Diophantine equation $1 + \sum_{j=1}^{n-1}\left(j \prod_{k=1}^j x_k\right) = \prod_{j=1}^n x_j$

Question

What are the positive solutions $(x_1,x_2,\ldots,x_n)$ for the Diophantine equation:

$$1 + \sum_{j=1}^{n-1}\left(j \prod_{k=1}^j x_k\right) = \prod_{j=1}^n x_j$$

Answer 1

It seems the following.

It can be easily checked by induction, that the equation has a simple solution $x_k=k$.

I investigated the equation for small values of $n$. It seems that its solutions can be obtained when we consequently consider admissible by divisibility values for $x_1,\dots,x_{n-1}$, branching our consideration. For instance, by this way we obtain

$n=1$. $1=x_1$

So $x_1=1$.

Solutions:

• $(1)$

$n=2$. $1+x_1=x_1x_2$.

So $x_1|1$ and $x_1=1$, $x_2=2$.

Solutions:

• $(1,2)$

$n=3$. $1+x_1+2x_1x_2=x_1x_2x_3$.

So $x_1|1$ and $x_1=1$. Then $2+2x_2=x_2x_3$.

So $x_2|2$ and $x_2=1$ or $x_2=2$.

Suppose that $x_2=1$. Then $x_3=4$.

Suppose that $x_2=2$. Then $x_3=3$.

Solutions:

• $(1,1,4)$
• $(1,2,3)$

$n=4$. $1+x_1+2x_1x_2+3x_1x_2x_3=x_1x_2x_3x_4$.

So $x_1|1$ and $x_1=1$. Then $2+2x_2+3x_2x_3=x_2x_3x_4$.

So $x_2|2$ and $x_2=1$ or $x_2=2$.

Suppose that $x_2=1$. Then $4+3x_3=x_3x_4$.

So $x_3|4$ and $x_3=1$ or $x_3=2$ or $x_3=4$.

Suppose that $x_3=1$. Then $x_4=7$.

Suppose that $x_3=2$. Then $x_4=5$.

Suppose that $x_3=4$. Then $x_4=4$.

Suppose that $x_2=2$. Then $3+3x_3=x_3x_4$.

So $x_3|3$ and $x_3=1$ or $x_3=3$.

Suppose that $x_3=1$. Then $x_4=6$.

Suppose that $x_3=3$. Then $x_4=4$.

Solutions:

• $(1,1,1,7)$
• $(1,1,2,5)$
• $(1,1,4,4)$
• $(1,2,1,3)$
• $(1,2,3,4)$

One may write a program which continues this consideration for larger $n$.