How to solve for $x$ in the following equation?

Question

Solve for $x \in \Bbb R$, given $$(x+m)^m-\binom{m}{1} (x+m-1)^m+\binom{m}{2}(x+m-2)^m + \cdots + (-1)^m \binom{m}{m} x^m=m!$$

Answer 1

This is an Identity! Therefore, this is true $ \forall x \in \mathbb R$. I came to know about this today, that's why I am sharing this with you all. Let's prove this :

Let's find coefficient of $y^m$ in $e^{xy}(e^y-1)^m$.

Expanding $(e^y-1)^m$ through binomial in LHS and Taylor series of $e^y$ in RHS, we've $$\text{Coefficient of } y^m \; \text{in} \; e^{xy}\left(e^{my}-\binom{m}{1} e^{(m-1)y}+\binom{m}{2} e^{(m-2)y}-\ldots (-1)^m\binom mm\right)=\text{Coefficient of } y^m \; \text{in} \; e^{xy} \left [ \left(1+y + \frac{y^2}{2!} +\ldots \right)-1\right]^m $$

Cancelling out $e^{xy}$ both the sides we get -

$$\boxed{\color{blue}{(x+m)^m-\binom{m}{1} (x+m-1)^m+\binom{m}{2}(x+m-2)^m \ldots (-1)^m \binom{m}{m} x^m=m!}}$$

Answer 2

Here is a slightly different variation on the proof that was given by OP. Introduce

$$P(x) = \sum_{q=0}^m (-1)^q {m\choose q} (x+m-q)^m.$$

We evaluate

$$[x^p] P(x)$$

where clearly $0\le p\le m.$ We have the claim if this is zero for $p\gt 0$ and $m!$ for $p=0.$

Extracting the coefficient yields

$$\sum_{q=0}^m (-1)^q {m\choose q} {m\choose p} (m-q)^{m-p} = {m\choose p} \sum_{q=0}^m (-1)^q {m\choose q} (m-q)^{m-p} \\ = {m\choose p} \sum_{q=0}^m (-1)^{m-q} {m\choose q} q^{m-p} \\ = {m\choose p} (m-p)! [z^{m-p}] \sum_{q=0}^m (-1)^{m-q} {m\choose q} \exp(qz) \\ = {m\choose p} (m-p)! [z^{m-p}] (\exp(z)-1)^m.$$

Note however that $\exp(z)-1 = z + \cdots$ and hence when $m-p\lt m$ we get a zero value because the coefficient vanishes. This condition is equivalent to $p\gt 0$ as required. On the other hand when $p=0$ we get

$${m\choose 0} m! [z^m] (\exp(z)-1)^m = m!$$

as claimed.