Sum of digits modulo a polynomial I made the following problems a while ago but I can't solve them (though I don't think it's too hard)

1.Let $s(n)$ be the digits sum of $n$. Let also $f(n)$, $g(n)$ $\in Z[X]$ . Assume that: 
  $$s(n) \equiv f(n) \pmod{g(n)} $$ for every $n$ (as long as $g(n)$ isn't $0$).
Then show that $g$ is a constant polynomial and $f$ has degree exactly $1$.

As I think you already observed the only congruences that  will work are derived from the classical $s(n) \equiv n \pmod{9}$ , for example :
$s(n) \equiv 19n-18 \pmod{9}$ 
What I tried :
Immediately I thought about the most important thing that connects polynomials and number theory : 
 $$a-b \mid g(a)-g(b)$$ 
Using this I got stuff like this :
$$g(n) \mid  kg(n) \mid g(n+kg(n))-g(n)$$ so 
$$g(n) \mid g(n+kg(n))$$
Now I made $k$ a super big power of $10$ bigger than $n$ ($k=10^l>n$)
So it's obvious that $s(n+kg(n))=n+g(n)$ for every such $k$ 
Now use the initial relation for two such $k$'s : $x$ and $y$ to get :
$$s(n+xg(n)) \equiv f(n+xg(n)) \pmod{g(n)}$$ and the analogue for $y$ which then subtracted gives :
$$f(n+xg(n)) \equiv f(n+yg(n)) \pmod{g(n)}$$ but this is already true so...I'm stuck.
Now the second problem :

2.Let $a$ be a positive integer and $g \in Z[X]$. Assume that:
  $$s(n) \equiv a^n \pmod{g(n)}.$$ Then show that $g$ is constant.

What I tried : I think that in this problem the previous method works better .
As before you can get :
$$a^{n+xg(n)} \equiv a^{n+yg(n)} \pmod{g(n)}$$ 
We can choose from the beginning an $n$ such that $g(n)$ and $a$ are coprime else it follows that $s(n)$ and $a$ are never coprime which is false for the $n$ made of $a+1$ ones .
So (assuming $x>y$ ) :
$$a^{(x-y)g(n)} \equiv 1 \pmod{g(n)}$$
This is where I'm stuck but I think that more can be done this way .
I think these problems are very interesting and (maybe) can be solved by only elementary means . I'd appreciate any thoughts .Thanks in advance for all the help.
 A: Here is a solution to question 1.
I am still thinking about question 2.
Solution to Question 1
First, we want to replace the polynomial $f$ with its remainder when divided by $g$. So we could write $f(x) = h(x) g(x) + r(x)$, where $\deg r < \deg g$.
Unfortunately, $h(x)$ and $r(x)$ could have rational coefficients rather than integer coefficients.
So instead, multiply out by their common denominator, and write
$$
M f(n) = h(n) g(n) + r(n)
$$
for some positive integer $M$, where $h, r$ are polynomials with integer coefficients, and $\deg r < \deg g$.
Now, fix an integer $k$, $k \ge 1$. Notice, there are infinitely many positive integers $n$ for which $s(n) = k$. So there are infinitely many $n$ such that $g(n) \mid f(n) - k$.
This implies $g(n) \mid M f(n) - Mk = h(n) g(n) + r(n) - Mk$, which implies $g(n) \mid r(n) - Mk$ for infinitely many $n$.
However, $\deg r < \deg g$.
If $\deg r \ge 1$, then $\deg g > \deg r$ implies that for sufficiently large $n$, $g(n) > r(n) - Mk > 0$, which contradicts that $g(n)$ divides $r(n) - Mk$ for infinitely many $n$.
So $r(n)$ is constant, say $r(n) = R \in \mathbb{Z}$.
In summary, there is a constant $R$ such that for any $k$, there are infinitely many values of $n$ such that $g(n)$ divides $R - Mk$.
But pick $k$ so that $R - Mk \ne 0$, and we get that $\boldsymbol{g(n)}$ must be a constant polynomial, as required.
Working further, let $g(n) = G \in \mathbb{Z}$.
Let $c$ be the constant coefficient of $f$.
For any prime $k \in \mathbb{Z}$, we have that $f(kp) \equiv c \pmod{G}$.
Therefore, as $G \mid f(n) - s(n)$ for all $n$,
$s(kG) \equiv c \pmod{G}$ for all $k$.
But pick $k = 10^j + 1$, where $j$ is large enough that $s(kG) = s(G) + s(G)$.
Then we get $s(G) + s(G) \equiv c \equiv s(G)$,
so $s(G) \equiv 0$, so $G \mid s(G)$.
But this implies $G \le s(G)$,
which can only happen if $G < 10$.
So $G \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
But we also have from $s(G) = G$ that $G \mid S(kG)$ for all $k \in \mathbb{G}$.
$G$ cannot be $2, 4, 5,$ or $8$,
because then $1000$ is a multiple of $G$, and $G \mid S(1000) = 1$.
Similarly, $G$ cannot be $6$, else $G \mid S(12) = 3$,
and $G$ cannot be $7$, else $G \mid S(14) = 5$.
So $\boldsymbol{g(n) = G \in \{1, 3, 9\}}$.
We are almost there.
Let $p(x) = f(x) - x$.
It is a familiar fact that for any of these three $G$,
$s(n) \equiv n \pmod{G}$ for all $n$.
So $f(n) \equiv s(n) \equiv n \pmod{G}$ for all $n$.
So $p(n) \equiv 0 \pmod{G}$ for all $n$.
Thus, all possible solutions are:
\begin{align*}
g(n) &= 1 \text{ and } f(n) \text{ is any polynomial;} \\
g(n) &= 3 \text{ and } f(n) = n + p(n), \text{ where } p(n) \text{ is a polynomial that is always a multiple of 3;} \\
g(n) &= 9 \text{ and } f(n) = n + p(n), \text{ where } p(n) \text{ is a polynomial that is always a multiple of 9.}
\end{align*}
As a side note, it is not true
that $f$ has degree exactly $1$, as you claimed.
Nor is it even true that e.g. in the case $g(n) = 9$, $f(n) = n + 9q(n)$ for some polynomial $q$.
For example, $f(n)$ could be $n + (n-1)(n-2)(n-3)\cdots (n-9)$.
The polynomial $p$ always returns a multiple of $9$ when evaluated,
but might not be a multiple of $9$ itself.
