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.