If $n \geq 3$, $\frac{m!}{k!}=n!-1$ so if $m-k \geq 2$, then $2 | n!-1$, and it's a contradiction. So $m-k =1$, and we get $(k,m,n)=(n!-2, n!-1, n)$ If $n=2$, $\frac{m!}{k!}=1$, so $(k,m,n)=(a, a, 2)$. If $n=1$, $m!=0$ so it's a contradiction. $\therefore (k,m,n)= (a, a, 2) \, (a \geq 1)$ or $(b!-2, b!-1, b) \, (b \geq 3)$.

Very easy but rather nice Diophantine here. Clearly, $k! + m! > k!$ so $n>1$. When $n=2$, we have \[k!+m!= 2!k!=2k!\]so $m=k$. Thus, we have the solution triples $(k,k,2)$ for all positive integers $k$. Next, if $n>2$, note that $k!$ divides the right hand side so $k! \mid k! + m! \implies k! \mid m!$. This means, $k\leq m$ and since $n>2$, $m>k$. Thus, \[k!n! = k! + m! = k!(1+ (k+1) \dots m)=k!(1+t)\]Now, if $m\geq k+2$, $(k+1)(k+2) \mid t$, and since one of these terms must be even, $t$ is even, implying that $t+1$ is odd. But, $n!$ is even for all $n>1$ which is a clear contradiction. Thus, $m=k+1$. Thus, we have the solution triples $(n! - 2 , n! -1 , n)$ for all positive integers $n>2$.

When $m<k$, $k!<k!+m!<2\cdot k!$, so there are no solutions for $n$. When $m=k$, $k!+k!=k!n!$ so $n=2$ and we get solutions $(k,k,2)$ for $k\ge1$. When $m=k+1$, $k!+(k+1)!=k!n!$ so $n!=k+2$ and we get solutions $(n!-2, n!-1, n)$ for $n\ge 3$. When $m>k+1$, $k!+m!=k!n!$ so $n!=1+\prod_{a=k+1}^ma>1$ and $n!$ is odd, so there are no solutions for $n$.

Notice that $n \geq 2$ and $\frac{m!}{k!} + 1 = n! \Longrightarrow$ $\frac{m!}{k!}$ is odd, then: $\bullet$ $m = k$ $\Longrightarrow n! = 2 \Longrightarrow$ $n = 2$, hence, $(k, m, n) = (t, t, 2)$ $\forall$ $t \in \mathbb{Z_+}$ $\bullet$ $m = k + 1$ ($k$ is even) $\Longrightarrow k + 2= n! \Longrightarrow$ $k = n! - 2$, hence, $(k, m, n) = (t! - 2, t! - 1, t)$ $\forall$ $t > 2 \in \mathbb{Z}$ $\therefore$ $(k, m, n) = (t, t, 2) \forall t \geq 1;$ $(t! - 2, t! - 1, t) \forall t \geq 3$

This is similar to BMO1 2002-3/5 except there was also an $n!$ on the LHS

Observe that $$1 + \frac{m!}{k!} = n!,$$and $n \ne 1$, so $ \frac{m!}{k! }$ is odd. In other words $m = k+1 \text{or} 0$, from where $(k,m,n)=(n!-2,n!-1,n), (k, k, 2)$ are the only solutions, which work.

$k!+m!=k!n!$ $\Rightarrow$ $1+\frac{m!}{k!}=n!$. So, $\frac{m!}{k!}=n!-1$. If $n\ge3$: Observe if $m-k\ge2$ then $2\mid\frac{m!}{k!}$ and $2\mid{n!-1}$, Contradiction!! as $n!-1$ is odd. So, ${m-k}\ngeq{2}$ $\rightarrow$ $m-k=1$ and $n\le2$. When $n=1$, $\frac{m!}{k!}=0$ but $ m!\neq{0}$, Contradiction. When $n=2$, $\frac{m!}{k!}=1$ $\Leftrightarrow$ $k=m$. Thus only satisfying integers $(k, m, n)=\boxed{x, x, 2}$ $\forall$ $x\ge1$$\in \mathbb{Z_+}$ $\bullet m=k+1$ $rightarrow$ $k!+(k+1)!=n!$ $\Rightarrow$ $k=n!-2$ and $m=n!-1$ $\forall{n\ge3}$ $\therefore$ $(k, m, n)=\boxed{x, x, 2}$ $\forall$ $x\ge1$$\in \mathbb{Z_+}$, $\boxed{n!-2, n!-1, n}$ $\forall$ ${n\ge3}$

The equation is equivalent to $$1 + \frac{m!}{n!} = n!.$$If $m \geq k+2$, the LHS is odd. It is also greater than 2, so this forces $n > 2$, implying that the RHS is even, a contradiction. We now consider the cases $m = k$ and $m = k+1$. $m=k$: we obtain $1 + \frac{m!}{k!} = 2 = n! \implies n = 2$. We obtain the solution $(k,k,2)$ for all $k \in \mathbb Z_+$. $m=k+1$: We get $k+2 = n! \implies k = n!-2$, $m = n!-1$, so the solutions are $(n!-2, n!-1, n)$ where $n \geq 3$ (since $k+2 \geq 3$). The solutions are $(k,k,2)$ for all $k \in \mathbb Z_+$ and $(n!-2, n!-1, n)$ where $n \geq 3$ (since $k+2 \geq 3$).

(k,m,n)=(a,a,2), (b!-2,b!-1,b) k=m or m-1 so, case 1)k=m->n=2 case 2)k=m-1->(m-1)!+m!=(m-1)!(m+1)

WOW! thdwlgh1229!!!!!

$v_2(k! + m!) = v_2(k!) + v_2(n!)$ note that if $v_2(m!) \neq v_2(m!)$, then $v_2(k! + m!) \leq v_2(k!) < v_2(k!) + v_2(n!)$, unless $n = 1$, which is obviously impossible note that obviously $m \geq k$ since $n > 1$. so $v_2(m!) = v_2(k!)$ just means $m = k + 1$ or $m = k$. so $k + 2 = n!$, and therefore $(n! - 2, n! - 1, n)$ is another solution!. so done. $m = k$ just means $2k! = n! \cdot k!$, which means $n = 2$, and $(k, k, 2)$ is another solution. done.

Let's respect RL_parkgong_0106 thdwlgh1229 KDH-2481632 YS_PARKGONG jaydenkaka jeusson

Case 1)$m=k$ $2k!=k!n!\implies n=2$ so $(k,m,n)=(k,k,2)$ Case 2)$m<k$ Then $k!n!=m!+k+<2k!\implies n!<2$ so $n=1\implies m!=0$ contradiction Case 3)$m>k$ If $m>k$ let $m=k+a$ plug in and we get: $k!+(k+a)!=k!n!$ $k!(1+(k+1)(k+2)\dots(k+a))=k!n!$ $1+(k+1)(k+2)\dots(k+a)=n!$ Case 3.1)$a>1$ If $a>1\implies (k+1)\dots(k+a)+1$ is odd, $n!$ is odd which means $n=1$ which is contradiction Case 3.2)$a=1\implies 2+k=n!$ which have infinite solution $(k,m,n)=(n!-2,n!-1,n)$ $\boxed{ANSWER:(k,m,n)=(k,k,2),(n!-2,n!-1,n)}$