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?

Here is an "almost" solution for integer coefficients.  
The zeros of $x^4+12x^3+14x^2-12x+1$ are algebraic integers.  So are their reciprocals.  Now in the series
$$
(1-xa)^{-1/4} = 1+\frac{a}{4} x+\frac{5 a^2}{32} x^2 +\frac{15a^3}{128} x^3 + \cdots
$$
where $a$ is an algebraic integer, the coefficients are algebraic integers (except for power-of-2 denominators).  Multiply four of these
$$
(x^4+12x^3+14x^2-12x+1)^{-1/4} = \prod_a (1-xa)^{-1/4}
$$
where the product is over all $a$ such that $1/a$ is a zero of $x^4+12x^3+14x^2-12x+1$.  Its coefficients are also algebraic integers, except possibly for power-of-2 denominators.  But this is invariant under permutation of the roots, so the coefficients are rational numbers.  (We can also see this more directly.)  Therefore the coefficients are rational integers, except possibly for power-of-2 denominators.  
After doing this, I thought I would be able to finish by using 2-adic numbers, to analyze these denominators.  But it seems this polynomial has no 2-adic zeros.  So I would have to go to an algebraic extension of the 2-adics...
