Determining all formal power series satisfying the equation $Q^m=(1+x)^n$ for $n,m\in\mathbb{Z}\setminus\{0\}$

Question

The question:

Let $m$ and $n$ be integers $\neq 0$. Determine all solutions $Q=1+b_1x+\cdots$ of the equation $$Q^m=(1+x)^n.$$

My current attempt at a solution:

Let $Q\in\mathbb{F}[[x]]$. The formal power series $1+x$ is equivalent to the binomial series $B_1$. Hence, we can rewrite the given equation in the following form \begin{align*} 1+\sum_{k=1}^\infty b_k^{(m)}x^k=Q^m=(1+x)^n=B_1^n=B_n=\sum_{k=0}^\infty\binom{n}{k}x^k. \end{align*} Applying the J.C.P Miller formula to $Q^m$ and comparing coefficients we construct the following linear system of equations \begin{gather*} b_0^{(m)}=b_0^m=1=1=\binom{1}{0}=[B_n]_0\\ b_k^{(m)}=\frac{1}{k}\sum_{l=1}^k[l(m+1)-k]b_lb_{k-l}^{(m)}=\binom{n}{k}=[B_n]_k \mbox{ for } k=1,2,\ldots. \end{gather*} Solving the second equation for $k=1$ we have, \begin{align*} \sum_{l=1}^1[l(m+1)-1]b_lb_{1-l}^{(m)}=mb_1b_0^{(m)}=mb_1=n=\binom{n}{1}. \end{align*} Hence, $b_1=\frac{n}{m}$. Solving the second equation for $k=2$ we have, \begin{align*} \frac{1}{2}\sum_{l=1}^2[l(m+1)-2]b_lb_{2-l}^{(m)}=\frac{1}{2}\left[(m-1)b_1b_1^{(m)}+2mb_2b_0^{(m)}\right]\\=\frac{1}{2}\left[\frac{(m-1)n^2}{m}+2mb_2\right]=\frac{n(n-1)}{2}=\binom{n}{2} \Rightarrow{} b_2=\frac{n(n-m)}{2m^2}. \end{align*} Hence, $b_2=\frac{n(n-m)}{2m^2}$. Suppose that the $k$th term of $Q$ satisfying the given equation is given by $b_k=\frac{\prod_{l=0}^{k-1}(n-lm)}{k!m^k}$ for $k\geq 1$. Then for $k+1$ it follows, \begin{align*} b_{k+1}=\frac{\prod_{l=0}^{k+1-1}(n-lm)}{(k+1)!m^{k+1}}=\frac{\prod_{l=1}^k(n-lm)}{(k+1)k!mm^k}=b_k\frac{n-km}{(k+1)m} \end{align*}

I can't seem to find a way to finish the induction step and have been unable to find an alternative method. Note that I am using $b_k^{(m)}$ to represent the $k$th coefficient of the $m$th power of the formal power series $Q$.

Edit: The field is considered to be an arbitrary field of characteristic zero.

Answer 1

You didn't specify what $\Bbb F$ was to be. Let me tell you what the answer is in case $\Bbb F$ is a field of characteristic $p>0$.

The easy case is the one where we’re asking for $m$-th roots of $(1+x)^n$ when $m$ is prime to $p$. Note that it’s enough to take the case when $n=1$. So we’re just asking for the existence of an $m$-th root of $1+x$, in $\Bbb F[[x]]$.

You could point out that the binomial coefficients in $(1+x)^{1/m}$ have denominators prime to $p$, but proving this seems a pain. Easier to observe that $1/m$ is a $p$-adic integer, i.e. is $p$-adic limit of a sequence of positive integers. In the $p$-adic topology, the powers of $p$ go to zero; and since the $p^k$ powers of $x$ go to zero in $\Bbb F[[x]]$ as $k\to\infty$, it follows that the $p^k$ powers of $1+x$, namely $(1+x)^{p^k}=1+x^{p^k}$ go to $1$ as $k\to\infty$.

Explicitly, then, let $\{\mu_0,\mu_1,\mu_2,\cdots\}$ be a sequence of positive integers with $1/m$ as $p$-adic limit. Then $$ (1+x)^{m_0},(1+x)^{m_1},(1+x)^{m_2},\cdots $$ is a convergent sequence in $\Bbb F[[x]]$, whose limit is an $m$-th root of $1+x$. Don’t see that right away? Well, what makes the sequence $\{m_i\}$ $p$-adically convergent is that $m_{i+1}\equiv m_i\pmod{p^{k_i}}$ for an unceasing sequence of integers $\{k_i\}$. That is, $m_{i+1}-m_i=p^{k_i}r_i$ for suitable integers $r_i$. Thus $$ \frac{(1+x)^{m_{i+1}}}{(1+x)^{m_i}}=(1+x)^{p^{k_i}r_i}=(1+x^{p^k})^{r_i}\longrightarrow1 $$

I’ll leave it to you to worry about the case where $p|m$.

Answer 2

I stepped away for a bit to see Lubin gave a much nicer answer. I can't guarantee there are no mistakes but my answer if probably an effectively more cumbersome version with more details left to fill in. Hopefully it does some good to post regardless!

We can use nonarchimedean analysis (ultrametric calculus) to put an absolute value on the elements of $\mathbb{F}[[X]]$ (and so get a metric).

$$|R|:= \begin{cases}a^{\mathrm{ord}(R)}& R \ne 0 \\ 0 & R = 0\end{cases}$$

Here $a$ is an element of $(0,1) \subset \mathbb{R}$ and $\mathrm{ord}$ is the order of the series $R$, the power on the lowest term (you can think of it as the opposite of the degree).

Here Hensel's lemma general form applies, for $f(Q) = Q^m-(1+X)^n$ we'd like to find a root. There is a solution if we can find some guess $Q_0$ such that,

$$|f(Q_0)|<|f'(Q_0)|^2$$

in our case we can simply pick $Q_0=1$ and we have (assuming the characteristic of our field does not divide $m$),

$$|1^m - (1+X)^n| \le |X| = a < 1 = |m*1^{m-1}|^2$$

Further the generalized Hensel lemma tells us this is the unique solution in the open ball of radius $|m*1^{m-1}|=1$ centered at $Q_0=1$, which means there is only one solution in $1 + X\mathbb{F}[[X]]$.

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At this point we can compute it by Newton's method, but we can do better and simply write the convergent series, as series converge nicely in ultrametric spaces - all you need is for their term to tend to $0$.

$$Q(z)=(1+X)^z = \sum_{k=0}^\infty \binom{z}{k} X^k$$

Here the binomial coefficient is defined as the image of the rational number in our field $\mathbb{F}$, $\binom{z}{k} = \frac{1}{k!} \prod_{i=0}^{n-1}(z-i)$

Because we have $Q(z) = (1+X)^z$ with $Q: \mathbb{N} \to \mathbb{F}[[X]]$ this infinite sum actually is finite, and so just a polynomial. We can extend it to rationals by putting a p-adic topology on $\mathbb{N}$ and slightly altering Mahler's theorem so the codomain is $\mathbb{F}[[X]]$ instead of $\mathbb{C}_p$. This gets us a unique continuous extension from $\mathbb{N}$ to the rational numbers without $p$ in the denominator. We simply pick $p$ so that it doesn't divide $m$ or the order of the field to avoid any issue.

Now let $z=\frac{n}{m}$, $$Q=(1+X)^{n/m} = \sum_{k=0}^\infty \binom{n/m}{k} X^k$$

For more detail I recommend looking at W. H. Schikhof's Ultrametric Calculus and Alain M. Robert's A Course in p-adic Analysis.