Basically the question is if number fields can be used
in [Aurifeuillean factorization](http://en.wikipedia.org/wiki/Aurifeuillean_factorization).


Probably this is easy and the answer is "no".


Let $f,g \in \mathbb{Z}[x], a \in \mathbb{N}$.
Let $f(x)$ and $f(g(x))$ be irreducible over $\mathbb{Q}[x]$.


The goal is to find nontrivial factor of $f(g(a))$ using number fields.


Suppose over some number field $K$ $f(g(x))$ is reducible and
factors $$f(g(x)) = f\_1(x) \cdot f\_2(x) \cdots f\_n(x)$$


The norms of $f\_i(a)$ will be related to the factorization
of $f(g(a))$, usually giving the trivial factor.



> 
> Is it possible for some $f,g,K$, the norm of $f\_i(a)$ to
>  give nontrivial factor of $f(g(a))$?
> 
> 
> 


I suspect the answer is "no".



Aurifeuillean factorization with number fields
I don't see the relevance of the extra polynomial $g$ in this question.


Anyway the answer is indeed no because if you let $\sigma\_1,\ldots,\sigma\_d$ be all different embeddings of $K \to \bar{\mathbb Q}$, then the norm of $f\_i(a)$ is $\prod\_j \sigma\_j(f\_i(a))$. However $f(g(x))$ is irreducible over $\mathbb Q$ so it has to divide $\prod\_j \sigma\_j(f\_i(x)) \in \mathbb Q(x)$. In particular the norm of $f\_i(a)$ will be a multiple of $f(g(a))$.


