# integer ordered pair of $(x,y,z)$ in $x!+y! = z!.\;,$ Where $x,y,z\in \mathbb{W}$

Total no. of integer ordered pair of $(x,y,z)$ in $x!+y! = z!.\;,$ Where $x,y,z\in \mathbb{W}$

$\bf{My\; Try::}$ Let $w=\max\left\{x,y\right\}$. Then $w<z$. So we can write $w\leq (z-1)$

So $w!\leq (z-1)!\Rightarrow w!\cdot z\leq z\cdot (z-1)!=z!=x!+y!\leq 2 w!$

So We get $z\cdot w!\leq 2w!\Rightarrow z\leq 2$

So If $z=2\;,$ Then we get $x=y=w$. So we get $2w!=2\Rightarrow w!=1\Rightarrow w= \left\{0,1\right\}$

So we get $(x,y,z) = \left\{0,0,2\right\}$ and $(x,y,z) = \left\{1,1,2\right\}$.

Similarly If $z=1$. Then $x!+y! = 1$. But $x!+y!\geq 2\forall x\in \mathbb{W}.$

Similarly If $z=0$. Then $x!+y! = 1$. But $x!+y!\geq 2\forall x\in \mathbb{W}.$

So We Get $(x,y,z)$ are $\left\{0,0,2\right\}$ and $\left\{1,1,2\right\}$

Is my solution is Right.

If Right can we solve it any other way.

Thanks

## 1 Answer

You are on the right track, but you did not consider all pairs $(x,y)$ where $x!=y!=1$. Also, you can be more concise, as follows.

Assume without loss of generality that $x\le y$. Then $z! = x!+y! \le 2y!$. Then $z!$ exceeds $y!$ by at most a factor of $2$. Since $x!$ is positive, $z!>y!$, and thus $z!$ must exceed $y!$ by more than a factor of $1$. Then $z!$ exceeds $y!$ by a factor of $2$ (as $y!$ divides $z!$). Then $y=0$ or $1$ and $z=2$. Then, removing the assumption $x\le y$, the possible solutions are $(0,0,2)$, $(0,1,2)$, $(1,0,2)$, and $(1,1,2)$.

Note that you don't need to introduce a new variable when you have two symmetric variables like $x$ and $y$ and you want to choose the greater one. Assume one is greater and, since they are symmetric, your proof will be valid under the opposite assumption as well (after the obvious changes).