# Is the set of roots of power series with integer coefficients countable?

Title says it all. It is clear that the set of all power series with integer coefficients is not countable while the set of algebraic numbers is countable, but I do not know whether the set of roots for power series with integer roots is uncountable.

Even ignoring the all-zeroes power series (of which every number is a root), it's still definitely uncountable. Given any $a\in (-1, 1)$ we can construct a nonzero power series $f_a$ with $a$ as a root. Basically, higher powers of $a$ yield smaller numbers, so we build $f_a$ as follows. First, suppose WLOG that $a\not=0$ (otherwise just make the constant term zero). Then note that (i) for all positive reals $\epsilon$, there is some even positive integer $n$ such that $0<a^{n}<\epsilon$, but (ii) for any positive $N$ and any positive even $n$ there is some positive integer $m$ such that $ma^n>N$.

This lets us recursively define a sequence of coefficients $a_i$ such that for each $n$, the sum $a_0+a_1a^2+a_2a^4+...+a_na^{2n}$ is between $0$ and ${1\over n}$; hence the whole power series goes to zero, and so $a$ is a root of $$\sum_{n\in\mathbb{N}}a_nx^{2n}.$$ Since there are uncountably many $x$ in $(-1, 1)$, this finishes the proof.

• That's what I get for typing on a cell phone. I didn't realize it was taking that long, and I didn't see your answer. Jun 3, 2017 at 20:31
• This is interesting, thanks. I wonder of there is a growth condition that can be placed on the coefficients. Similarly, can you get the same result if you bpundits the coefficients. If the answer is not trivial, I can ask separately.
– Paul
Jun 3, 2017 at 22:03
• @MattSamuel Well, your answer goes into much more detail than mine. Jun 3, 2017 at 22:11

Let $0<a<1$ be irrational. I claim that there is a power series $$\sum{a_nx^n}$$ with integer coefficients that vanishes at $a$.

If $a>\frac12$, let $a_0=1$ and $a_1=-1$, otherwise let $a_0=0$ and $a_1=1$. Suppose recursively that $k\ge 1$ we have defined $a_0,\ldots,a_N$ such that $$0<\sum_{n=1}^N{a_na^n}<\frac1{k+1}$$ We choose $N'$ and $a_{N+1},\ldots,a_{N'}$ such that $$0<\sum_{n=1}^{N'}{a_na^n}<\frac1{k+2}$$ If the sum (let's call it $A$ for brevity) already satisfies this let $N'=N$. Otherwise choose $N'$ large enough that $$a^{N'}<\frac1{2(k+2)}$$ Let $a_i=0$ if $N<i<N'$. Then let $$a_{N'}=-\left\lceil\frac{(A-\frac1{k+2})}{a^{N'}}\right\rceil$$ Then $$0<A+a_{N'}a^{N'}<\frac1{k+2}$$ Letting $k\to\infty$, we end up with a power series with integer coefficients that vanishes at $a$, which should answer your question.