Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients?  (I have others to do, too, but this one will be a start.)
possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
(b) $\mu(x)$ satisfies the differential equation
$$
(x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0
$$
(c) factorization of $x^4+12x^3+14x^2-12x+1$ is
$$
\left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right)  \left( x-\sqrt 
{5}+3-\sqrt {15-6\,\sqrt {5}} \right)  \left( x+\sqrt {5}+3-\sqrt {15+
6\,\sqrt {5}} \right)  \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}}
 \right) 
$$
(d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.
(e) Can we compute the series for $\log\mu(x)$
$$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{
\frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots
$$
and then recognize that its exponential has integer coefficients?

For the periodicity mod $5$, note that 
$$(1 - x^4)^4 \equiv (1 + 3 x + 4 x^2 + 2 x^3)^4 (1-12 x+14 x^2+12 x^3+x^4) \mod 5 $$
and thus
in the field $\mathbb Z_5((x))$ of formal Laurent series in $x$ over the integers mod $5$ we have
$$ \eqalign{&\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \left(\dfrac{1+3x+4x^2+2x^3}{1-x^4}\right)^4\cr 
&= \left(1+3x+4x^2+2x^3 + (1+3x+4x^2+2x^3) x^4 + (1+3x+4x^2+2x^3) x^8 + \ldots  \right)^4\cr} $$
Given there is a solution over the integers, taking the coefficients mod $5$ gives us a solution in $\mathbb Z_5((x))$.  Of course there are four fourth roots, but this is the only solution whose constant coefficient is $1$ (the others have $2, 3$ or $4$).
Similarly, to show the coefficients are odd, 
$$(1-x)^4 \equiv 1 + x \equiv 1 -12 x + 14 x^2 + 12 x^3 + x^4 \mod 2$$
so that in $\mathbb Z_2((x))$
$$\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \dfrac{1}{(1-x)^4} = (1 + x + x^2 + x^3 + \ldots)^{4}$$
Combine the solutions mod $5$ and mod $2$ and we see that mod $10$ the coefficients repeat $1,3,9,7$.
The pattern mod $3$ also seems quite interesting (but not periodic): I'll leave that as an exercise.  Solution tomorrow if nobody else posts it.
EDIT: OK, time's up.  Mod $3$ we have 
$$\eqalign{(1 - x^2 + x^4)^{-1/4} &\equiv 1+{x}^{2}+{x}^{6}+{x}^{8}+{x}^{18}+{x}^{20}+{x}^{24}+{x}^{26}
\ldots\cr
&= \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right) = \sum_{k \in S} x^{k}\cr}$$
where $S$ is the set of positive integers whose base-3 representation contains no $1$'s.  This is because
$$(1+x^{m})^3 \equiv 1 + x^{3m}$$ so that
$$ \dfrac{1+x^{2\cdot 3^{n+1}}}{1+x^2} \equiv \prod_{j=0}^n \dfrac{1+x^{2\cdot 3^{j+1}}}{1+x^{2\cdot 3^j}} \equiv \prod_{j=0}^n \left(1 + x^{2\cdot 3^j}\right)^2$$
Taking $n \to \infty$ we get
$$ \dfrac{1}{1+x^2} \equiv \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right)^2$$
and then 
$$ \dfrac{1}{1-x^2 + x^4} \equiv \dfrac{1}{(1+x^2)^2} \equiv 
\prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right)^4 $$
