# Integer solutions of $x! = y! + z!$

There was an interesting problem asked about triples $(x,y,z)$ which are solutions of

$$x! = y! + z!.$$

Here $(2,1,1)$ is a solution because $2! = 1! + 1!$, as are $(2,1,0)$ and $(2,0,1)$.

Now I wanted to analyze this a bit further and thought of using the gamma function definition of a factorial, to see where it led and this is what I got:

$x! = y! + z!$

$\Gamma(x) = \Gamma(y) + \Gamma(z)$

$\int_{-\infty}^{\infty}t^xe^{-t}dt = \int_{-\infty}^{\infty}t^ye^{-t}dt + \int_{-\infty}^{\infty}t^ze^{-t}dt$

$\int_{-\infty}^{\infty}t^xe^{-t}dt - \int_{-\infty}^{\infty}t^ye^{-t}dt - \int_{-\infty}^{\infty}t^ze^{-t}dt = 0$

$\int_{-\infty}^{\infty}(t^x - t^y - t^z)e^{-t}dt = 0$

Now my line of thinking was that, in a manner similar to the fundamental lemma of the calculus of variations, this should imply that the above integral can only be true for arbitrary values of $x, y$ and $z$ if $t^x - t^y - t^z = 0$.

I'm not really sure if that is justifiable so I'd appreciate some comment on this.

The reason I continued on despite the uncertainty is that when you're left with the polynomial $t^x - t^y - t^z = 0$ you encounter a strange fact. First off, plugging in a triple like $(2,1,1)$ results in $t^2 - t^1 - t^1 = 0$ which implies $t^2 = 2t$ thus $t = 2$. Now plugging in $t = 2$ gives $2^2 = 1^2 + 1^2$. In other words you get what you'd expect. However plugging in a triple you know is inconsistent, like $(0,0,0)$, gives

$t^0 = t^0 + t^0$

$1 = 1 + 1$

$1 = 2$.

You get absurdities when you plug in inconsistent triples.

So while $(2,1,0)$, which works, gives

$t^2 - t^1 - t^0 = 0$

$t^2 - t - 1 = 0$

whose roots are

$\frac{1}{2} + \frac{\sqrt{5}}{2}$

$\frac{1}{2} - \frac{\sqrt{5}}{2}$

and on plugging these into $t^x - t^y - t^z = 0$ gives a consistent equality for both roots, an inconsistent triple like $(3,2,1)$ leaves us with $t^3 - t^2 - t = 0$, one of whose roots are $t = 0$ (which gives a consistent equality $0^3 - 0^2 - 0 = 0$ whereas another one of it's roots, $t = \frac{1}{2} + \frac{\sqrt{5}}{2}$ does not give a consistent equality

$(\frac{1}{2} + \frac{\sqrt{5}}{2})^3 - (\frac{1}{2} + \frac{\sqrt{5}}{2})^2 - (\frac{1}{2} + \frac{\sqrt{5}}{2}) = 0$.

If all of the above is magically okay you can use this idea to show that no triples $(n, n - 1, n - 2)$ will work for $n$ greater than $2$. I'd like to go further with it but I'm afraid it's all just conceptually flawed, is it?

A more interesting question stems from something a great lecturer told me, she basically analyzed the whole question geometrically and said something about certain triples having to do with the area of those gamma integrals canceling out. I thought I understood what she was saying but I actually don't so I'd appreciate any comment on what this could mean.

I hope I was clear enough, thanks for your time.

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The reasoning after and including calculus of variations doesn't make sense to me. Also I would note: this isn't the usual definition of the gamma function, which interpolates as $\Gamma(x) = (x-1)!$. –  Cocopuffs Oct 3 '12 at 17:23
Wow, I'm such an idiot. Based on forgetting the - 1 term I went in circles doing a lot of nonsense. Just delete this post, no marks, complete & utter failure... Thanks for your time. –  sponsoredwalk Oct 3 '12 at 18:13
no! upvote for you for forging ahead in muddy waters! –  im so confused Oct 3 '12 at 20:37

$x! = y! + z!$ does not have any solutions in integers with $x \geq 3.$ As soon as $x \geq 3,$ we have $(x-1)! \leq x! / 3.$ With the necessary $y,z < x,$ we get $y! \leq x! / 3, \; \; z! \leq x! / 3,$ so $y! + z! \leq 2 x! / 3$ and $y! + z! \neq x!$

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Neat solution. You can rewrite the solution this way: $y,z <x$ hence $y,z \leq x-1$, and thus $x!=y!+z! \leq (x-1)!+(x-1)!$ which implies $x \leq 2$. –  N. S. Oct 3 '12 at 17:46
The more interesting problem is $x! = y! \; z!.$ For any $n,$ we know $n! = (n-1)! \; n.$ Taking $n = z!,$ we get the infinite family $(z!)! = (z! - 1)! \; z!.$ I think there is at least one other solution, the question is to find others. It's on MO somewhere. –  Will Jagy Oct 3 '12 at 18:07
@Will. It's here too. –  Rick Decker Oct 3 '12 at 18:21
@RickDecker thanks, I think that is actually where I saw it. Also I had not remembered $6! \; 7! = 10!$ –  Will Jagy Oct 3 '12 at 18:34