NT from Vietnam 2005

[mutasimmim]

Find all nonnegtive integer solutions to $\dfrac{x!+y!}{n!}=3^n$

[Fm Jakaria]

$n = 0$ is impossible. So let for advantage, $x,y,n > 0$. Assume $y \geq x$.

Now $x!(1+ \frac{y!}{x!}) = n!3^n…..(1) $

First case:
First let $x \geq n$. Then $d = \frac{x!}{n!} (1+ \frac{y!}{x!}) = 3^n$ is odd. So $n+1 \geq x$ and $y > x$.
Subcase 1: If $x = n+1; (n+1)(1+ \frac{y!}{(n+1)!}) = 3^n$. So $y \leq n+3$.
For $ y = n+2; (n+1)(n+3) = 3^n; (n+1)$ and $(n+3)$ are both odd and positive powers of three. So they are relatively prime;
impossible as $n+1 > 1$.
Now let $y = n+3, and n + 1 = 3^k$ for some positive integer $k < n$. Then $3 + 3^{2k} + 3^{k+1} = 3^{n – k}$. So $3$ fully divides LHS, so RHS is $3$; but LHS $>3$; a contradiction.

Subcase 2:
Now let $x = n$, ($1+ \frac{y!}{x!}) = 3^x$; so $x+2 \geq y$. So $1+(x+1)(x+2) = x^2 + 3x +3 \geq 3^x$.
Note that if for integer $a > 0$, $a^2 + 3a + 3 < 3^a$, then $(a+1)^2+3(a+1)+3 < 3a^2+9a+9 < 3^{a+1}$. So $x < 3$. Note $x = 2$ doesn’t work.
$x = 1$ gives $y = 2$.

Case-2:
Now let $x<n$. Then (1) gives $c = 1+ \frac{y!}{x!} = \frac{n!}{x!} 3^n$. So $(x+1) (>1) | c$. So $x = y$. Then
$x+1 (> 1)| c = 2$ implies $x = 1$. Then $n! (>1) |2$ imples $n = 2$. Then $3^n = 1$, absurd.

So only solutions are: $(n,x,y) = (1, t, 2) , (1, 2, t)$ for $t \in $ {$0,1$}.

[mutasimmim]

n=0 is impossible.

Why?

$x,y,n>0$

Again why?

[Fm Jakaria]

Well, for the first question; a factorial(such as $x!$ and $y!$) is always $\geq 1$. So $n$ can't be $0$.
For the second question, I haven't said $x,y,n>0$. Here $0$ and $1$ has same factorial $1$; so I have said
"So let for advantage, $x;y;n>0$.