# Finding all the numbers that fit $x! + y! = z!$

I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it.

From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$

So we can do something like this:

$$[x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$

we can then factor all of the common terms out on the LHS. $$[...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$ and divide the common terms out of the right hand side $$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$

I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution).

If anybody can provide a hint as to how to proceed I would appreciate it.

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Suppose without loss of generality that $x \le y$. Then $z!\le 2y!$. – Chris Eagle Feb 5 '12 at 0:23
Does Modular Arithmetic help? – user21436 Feb 5 '12 at 0:35
– Martin Sleziak Aug 14 '14 at 4:29

If $x, y \in \{0,1\}$, then we can always find a solution $z \in \{0, 1, 2\}$. The rest of the post will show that there are no other solutions.

Let us assume $y \geq x \geq 2$ without loss of generality.

Dividing both sides by $x!$ gives $$1 + y(y-1)\cdots(x+1) = z(z-1)\cdots(x+1).$$ If $y > x$, we see $x+1$ divides the right-hand side but not the left-hand side ($x+1$ divides one term in the sum but not the other), in which case there are no solutions.

If $y = x$, we may reduce the problem to that of solving $2y! = z!$. Since $y \geq 2$, the left-hand side always has more factors of $2$ than the right-hand side, in which case there are no solutions.

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If $x>1$ and $y \leq x$ then $$x!<x!+y! \leq 2x!<(x+1)!$$ therefore $$x!<z!<(x+1)!$$ so that the only solutions are $(x,y,z)=(0,0,2),(0,1,2),(1,0,2),(1,1,2)$.

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How does the $z!$ come into play in your second equation? – Roland Feb 15 at 19:42
We already have $z! = x! + y!$ – jelec Feb 15 at 19:49

Very shortly, $(x,y,z)=(0,0,2),(0,1,2),(1,0,2),(1,1,2)$ are the only solutions because if WLOG $1<x\le y<z$ then dividing by $y!$ yields $$1+\frac{1}{(x+1)(x+2)\cdots y}=(y+1)(y+2)\cdots z,$$ whose RHS is an integer while its LHS is not.

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The problem is how to show it

One way maybe using Stirling's formula that approximates $n!$ as follows:

$n! \approx n \ln (n) - n$

so you could write:

$x \ln(x) - x + y \ln(y) - y \approx z \ln(z) - z$

$z - x - y \approx z \ln(z) - x \ln (x) - y \ln (y)$

one solution to this may be derived by:

$z=z\ln(z)$ and $x=x\ln(x)$ and $y=y\ln(y)$

that is:

$1=\ln(z)$ and $1 = \ln(x)$ and $1 = \ln(y)$

this leads us to the fact that $x, y, z$ are all between $0$ and $e+m$ where $m$ is a small integer greater than or equal to zero. I used the $m$ here since the Sterling formula is not accurate hence the values may not be exact. One could then try manually integers in the range $[0,2+m]$ and construct the different combinations to find at least $1$ solution.

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Unfortunately for the assignment we've been instructed to stay in the integers. So we can't use ln() and such. – AvatarOfChronos Feb 5 '12 at 0:56
A harder problem is x! y! = z!. – marty cohen Feb 5 '12 at 3:21
@Brian M. Scott - Thanks for editing! – NoChance Feb 5 '12 at 6:56

If $x!+y!=z!$ in positive integers $x,y,z$, we can assume $x\le y$, and it's clear that $y\le z$. These inequalities imply $x!\mid y!$ and $y!\mid z!$. But

$$y!\mid z!\implies y!\mid(z!-y!)\implies y!\mid x!$$

and this implies $y=x$, so that $z!=2x!$. The only solution to this is $x=1$, $z=2$, so we get $(x,y,z)=(1,1,2)$ as the only solution in positive integers.

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