About a subset of $\mathbb Q[x]$ of polynomials $f$ such that $f(n)=f(-n)$ for every $n$ in $\mathbb N$ Let $A=\{f \in \mathbb Q[x] : f(n)=f(-n)$, for every $n \in \mathbb N\}$.
Show that

*

*$A$ is a subring of $\mathbb Q[x]$.

*$A$ is a Euclidean Domain.

*For every $f \in A$ we have $f(r)=f(-r)$, for every $r \in \mathbb Q$.

I think I managed to prove 1 and 2, this is how I proceeded:
1
We have that $A \neq \emptyset$, since the constant polynomial $1$ is such that $1(n)=1=1(-n)$, for every $n \in \mathbb N$.
For every $f$, $g \in A$, for every $n \in \mathbb N$ we have:
$(f-g)(n)=f(n)-g(n)=f(-n)-g(-n)=(f-g)(-n)$
and
$(fg)(n)=f(n)g(n)=f(-n)g(-n)=(fg)(-n)$
These mean that both $f-g$ and $fg$ are in $A$. This proves that $A$ is a subring of $\mathbb Q[x]$.
2
For every $f \in A$, we have:
$f=a_nx^n+\cdots+a_0$
where, for each $i \in \{0,\ldots,n\}$, $a_i \in \mathbb Q$.
We have that, for each $q \in \mathbb Q$ seen as a polynomial in $\mathbb Q[x]$, $q(n)=q=q(-n)$, for every $n \in \mathbb N$. This implies that $\mathbb Q \subseteq A$.
If every rational number is in $A$, this implies that every polynomial $f$ in $A$ has a leading coefficient $a_n$ that is a unit in $A$.
I consider the function $deg : A \to \mathbb N$, which assigns to every $f \in A$ its degree. This function serves as a Euclidean evaluation, that means it satisfies:
For every $f$, $g \in A$, with $g \neq 0$, there exist $q$, $r \in A$ such that:

*

*$f=gq+r$

*$r=0$ or $deg(r)<deg(g)$
This evaluation, together with the fact that every polynomial in $A$ has a unit as leading coefficient, ensure that $A$ is a Euclidean domain.
Now, my questions are:

*

*Are my solutions correct?

*Can you help me with point 3? There should be some way to exploit the fact that every pair of polynomials can go through Euclidean division but I can't see where to begin.


If $f(x)\in A$, let $g(x)=f(x)-f(-x)$. Then $(\forall n\in\Bbb N):g(n)=0$. Since $g(x)$ has infinitely many zeros, it is the null polynomial. Therefore, $(\forall r\in\Bbb Q):f(r)=f(-r)$.
