# Math Help - factorial equation

1. ## factorial equation

Solve $x!+y!=z!$ in the set of non-negative integers.

It seems that $0!+1!=2!$ would work but I'm not totally sure. Is there any "official" solution or is guesswork the only way to go?

2. I believe this should be in the Number Theory Section.

You might want to consider the following expression which is equivalent :

$2 \times 3 \times \cdots \times x + 2 \times 3 \times \cdots \times y = 2 \times 3 \times \cdots \times z$

You can factorize this, then divide the RHS to the number-only LHS factor. Then try to see when it actually is an integer ()

EDIT : this might not give all solutions. You can also study the equation $x! - z! = y!$ which is conceptually simpler.

3. Assume without loss of generality that $x\le{y}$, and divide by x!:

$1+\frac{y!}{x!}=\frac{z!}{x!}$

Now, what is the relationship between y! and x!?

You should get that there are 3 solutions besides the one you gave.

4. Hello,

If x=y, then we'd have 2*x!=z!
But unless x=1, it's not possible to have a solution.
So (1,1,2) is a solution.

If x<y (wlog), we can divide each side by x! and we'd have $1+(x+1)\cdots (y-1)y=(x+1)\cdots (z-1)z$

It's obvious that y<z. So the RHS can be written $(x+1)\cdots(y-1)y(y+1)\cdot (z-1)z$

Hence $1=(x+1)\cdots(y-1)y\cdot[z(z-1)\cdots (y+1)-1]$

Since all of these are integers, we must have $(x+1)\cdots(y-1)y=1$ and $z(z-1)\cdots (y+1)-1=1$

Again, since all the factors are integers, we must have $x+1=y=1 \Rightarrow x=0,y=1$

Thus there are only 2 solutions : (0,1,2) and (1,1,2)

5. I was thinking of (0,0,2) and (1,0,2) in addition to your solutions (0,1,2) and (1,1,2).