# Problem 9.4

Solve the equation in natural numbers $\mathrm{n}$ and $\mathrm{m}$

$(n+1)!(m+1)!=(n+m)!$

Number of points 7
Answer:

$\{(2 ; 4),(4 ; 2)\}$

## Solution

Obviously, $\mathrm{n}>1$ and $\mathrm{m}>1$ (since when $\mathrm{n}=1$ we get $2(m+1)!=(1+m)!$ and similarly when $\mathrm{m}=1$).

Divide the equation by $(n+1)!$, we get $1 \cdot 2 \cdot 3 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$

Here on the left side there are $\mathrm{m}+1$ factors, on the right side there are $\mathrm{m}-1$.

When $\mathrm{n}>3$ we get $(1 \cdot 2 \cdot 3) \cdot 4 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$

Or $6 \cdot 4 \cdot \ldots \cdot(m+1)=(n+2)(n+3) \ldots(n+m)$

Here we have the same number of factors, and when $\mathrm{n}>3$ we have

$6 \leq n+2$

$4<n+3$

$5<n+4$

$\cdots$

$(m+1)<(n+m)$

Therefore, the equality is impossible. Thus, if (n, m) is a solution, then $1<\mathrm{n}<4$. By the symmetry of the equation, if ($\mathrm{n}, \mathrm{m}$) is a solution, then $1<\mathrm{m}<4$. (These statements are not necessarily true simultaneously).

Now let $\mathrm{n}=2$, then $(1 \cdot 2 \cdot 3) \cdot(1 \cdot 2 \cdot \ldots \cdot(m+1))=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(2+m)$

Or after simplification $(1 \cdot 2 \cdot 3)=(2+m)$

From this, $\mathrm{m}=4$

Now let $\mathrm{n}=3$, then $(1 \cdot 2 \cdot 3 \cdot 4) \cdot(1 \cdot 2 \cdot \ldots \cdot(m+1))=1 \cdot 2 \cdot 3 \cdot \ldots \cdot(3+m)$

Or after simplification $(1 \cdot 2 \cdot 3 \cdot 4)=(2+m)(3+m)$

There are no natural solutions