Find all solutions of the equation $x! + y! = z!$

Question

Not sure where to start with this one. Do we look at two cases where $x<y$ and where $x>y$ and then show that the smaller number will have the same values of the greater? What do you think?

Answer 1

Note that if $z$ is $3$ or more, there will be no solution as $2(z-1)! \lt z!$. That doesn't leave many possibilities to check.

Answer 2

Let $x \le y < z$:

We get $x! = 1\cdot 2 \cdots x$, $y! = 1\cdot 2 \cdots y$, and $z! = 1\cdot 2 \cdots z$.

Now we can divide by $x!$ $$1 + (x+1)\cdot (x+2)\cdots y = (x+1)\cdot (x+2) \cdots z$$

You can easily show by induction that the right side is bigger than the left for all $z>2$. The only cases that remain are $0! + 0! = 2!$, $0! + 1! = 2!$, $1! + 0! = 2!$, and $1! + 1! = 2!$.

Answer 3

Assume that $x\ne y$, say $x\lt y$, then one knows that $x\leqslant y-1$ and that $z\gt y$ hence $z\geqslant y+1$. Thus, $z!\geqslant y!(y+1)$ and $x!+y!\leqslant y!(1+1/y)$. If these are equal, $1+1/y\geqslant y+1$. Can you finish this case and find a similar argument when $x=y$?

Answer 4

Suppose without loss of generality that $x\leq y$. We will necessarily have to have $y\leq z$, as well.

If $x=y$, then $x!+y!=2\cdot x!$, which can only be a factorial if $x=0$ or $x=1$. If $x<y$, then dividing by $x!$ yields $$1+y\cdot(y-1)\cdots(x+1)=z\cdot(z-1)\cdots(x+1).$$ We then clearly can't have $z=y$ (why?), so after subtraction and factoring, we have $$1=\bigl[z\cdots(y+1)-1\bigr]\cdot y\cdot(y-1)\cdots(x+1).$$ This only works if $$z\cdots(y+1)=2$$ and $$y\cdot(y-1)\cdots(x+1)=1.$$ (why?) This implies that $y=1=x+1$ and $z=2$.

Thus, you only need to consider triples $(x,y,z)$ of non-negative integers with $z=2$.

Answer 5

Suppose that $0 \le x \le y$ WLOG. Then $y! < x! + y! \le 2(y!) < (y+1)!$ for $y \ge 2$, so no solutions with $y \ge 2$ are possible (because then every $x! + y!$ is between two consecutive factorials and not equal to either one).