Power series representing a rational function This question is related to lemma II on page 4 of this book by R. Salem on Fourier Analysis and Number Theory. For completeness, I give the statement of the lemma, and a link to the proof: 

If in the series $f(z)=\sum_{n=0}^\infty c_nz^n$ the coefficients $c_n$ are integers, and if the series represents a rational function, then
  $$f(z)=\frac{P(z)}{Q(z)},$$
  where $P/Q$ is irreducible, $P$ and $Q$ are polynomials with integer coefficients, and $Q(0)=1$. 

The proof can be found here (I've omitted the first part, where he shows that a product of primitive power series is primitive). 
In the last part of the proof, he writes

Now let $U$ and $V$ be polynomials with integral rational coefficients such that 
  $$PU+QV=m\neq 0,$$
  $m$ being and integer. Then
  $$m=Q(Uf+V).$$
  Since $Q$ is primitive, $Uf+V$ cannot be primitive, for $m$ is not primitive unless $|m|=1$ Hence, the coefficients of $Uf+V$ are divisible by $m$.

My question is: Why does the last sentence follow? I'm guessing it has something to do with the fact that $Uf+V$ is primitive, and I've tried several things but I can't see it. Also, the fact that he calls this result "Fatou's lemma" makes finding references quite hard (I couldn't find any). 

I guess the idea is the following. See here that the product of two polynomials is primitive if and only if both polynomials are primitive. Write 
$1=Q \quad \frac{Uf+V}{m}$
$1$ is  primitive, and it is a product of two polynomials. $Q$ is primitive by construction, hence $\frac{Uf+V}{m}$ must also be primitive, hence the coefficients of $Uf+V$  must be divisible by m.
