Solve $\displaystyle x!+y!=z!$ in the set of non-negative integers.
It seems that $\displaystyle 0!+1!=2!$ would work but I'm not totally sure. Is there any "official" solution or is guesswork the only way to go?
I believe this should be in the Number Theory Section.
You might want to consider the following expression which is equivalent :
$\displaystyle 2 \times 3 \times \cdots \times x + 2 \times 3 \times \cdots \times y = 2 \times 3 \times \cdots \times z$
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 $\displaystyle x! - z! = y!$ which is conceptually simpler.
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 $\displaystyle 1+(x+1)\cdots (y-1)y=(x+1)\cdots (z-1)z$
It's obvious that y<z. So the RHS can be written $\displaystyle (x+1)\cdots(y-1)y(y+1)\cdot (z-1)z$
Hence $\displaystyle 1=(x+1)\cdots(y-1)y\cdot[z(z-1)\cdots (y+1)-1]$
Since all of these are integers, we must have $\displaystyle (x+1)\cdots(y-1)y=1$ and $\displaystyle z(z-1)\cdots (y+1)-1=1$
Again, since all the factors are integers, we must have $\displaystyle x+1=y=1 \Rightarrow x=0,y=1$
Thus there are only 2 solutions : (0,1,2) and (1,1,2)