amparvardi wrote: Does there exist an integer $z$ that can be written in two different ways as $z = x! + y!$, where $x, y$ are natural numbers with $x \le y$ ?

Let $a! + b! = z = x!+ y!$ and $a > x \geq y > b$=> $a!- x! - y!+b!=0 $ ($ \nu_p (m)$ - such $k$ that $p^k | m$ and $p^{k+1}$ doesn't divide $m$) If $y \geq b+2$ then $ \nu_2 (a!) \geq \nu_2 (x!) \geq \nu_2 (y!) > \nu_2 (b!)$ => $a!, x!, y!$ divides by $ \nu_2 (y!)$ and $b!$ doesn't divided by it- contradiction. So $y=b+1$ (because $b+2>y>b$). Then $a!-x! \geq (x+1)!-x! = x(x!) > b(b!) = (b+1)! - b! = y! - b!$- contradiction. So such $z$ don't exist

We can check that $z\geq 6$. Let $n!\leq z < (n+1)!$ for some integer $n\geq 3$. If $y<n$, then $x!+y!\leq 2\cdot (n-1)!<n!\leq z$, impossible. Also, $y<n+1\implies y=n$, which forces $x$, if possible, and thus the answer is no.

The answer is $\boxed{\text{no}}$. Let $z=x_1!+y_1!=x_2!+y_2!$, where $x_1\leq x_2\leq y_2\leq y_1$; we will show that $x_1=x_2$ and $y_1=y_2$. Here, the $x_i\leq y_i$ are in accordance with the format given in the problem, and $x_1\leq x_2$ is assumed without loss of generality. As a result of this assumption, $x_2,y_2\leq y_1$, since otherwise $y_1<y_2\implies y_1!<y_2!$, and summing with $x_1\leq x_2\implies x_1!\leq x_2!$ yields $x_1!+y_1!<x_2!+y_2!$, contradiction. Now, in the equation $x_1!+y_1!=x_2!+y_2!$, take $\pmod{x_2!}$; then, since $x_2,y_1,y_2\geq x_2$, we have that $x_2!,y_1!,y_2!\equiv0\pmod{x_2!}$, which implies that $x_1!\equiv0\pmod{x_2!}$, i.e, $x_1\geq x_2$. But by assumption $x_1\leq x_2$, so $x_1=x_2$, and substituting into $x_1!+y_1!=x_2!+y_2!$ gives $y_1=y_2$, which together suffice. $\blacksquare$

Assume that there is such an integer $z$. Let $z=a!+d!=b!+c!$, where $a < c \leq d < b$. Note that $b \geq 3d! \geq 3c! \Rightarrow c!+d! <b!$, which is a contradiction. Therefore, there is no such $z$.