Ideal of $\mathbb{Z}[x]$ generated by $3$ and $x^2+1$ is proper 
Let $I$ be the ideal of $\mathbb{Z}[x]$ generated by $3$ and $x^2+1$. Show that
1) $I$ is proper,
2) $\phi_a(I)=\mathbb{Z}$ for all $a\in\mathbb{Z}$, where $\phi_a(I):=\{f(a)\mid f\in I\}$ (evaluation).

My attempt: 1) $1\not\in I$. Otherwise, we should be able to find the integers $i_k, j_k$ such that $i_kj_k\not=0$ and $1 = h(x)\cdot\sum_k 3^{i_k}(x^2+1)^{j_k}$ for some integer $k>0$ and for some polynomial $h(x)\in\mathbb{Z}[x]$, since $I$ is generated by $3$ and $x^2+1$. So $j_k=0$, $h(x)=n$: constant.  and we have nonzero $i_k,n$ : $1\not=n\cdot\sum_k 3^{i_k}$. Thus $1\not\in I$, and $I\not=\mathbb{Z}[x]$.
2) If we find $f$ such that $f(a)=1$, then $\phi_a(I)=\mathbb{Z}$ for that $a$ (because $I$ is an ideal of $\mathbb{Z}[x]$, and $\mathbb{Z}\subset \mathbb{Z}[x]$, $n\cdot f\in I\implies n\cdot f(a)\in\phi_a(I)$ for all $n\in \mathbb{Z}$). So the question is reduced to find $f$ generated by $3$ and $x^2+1$ such that $f(a)=1$ for each $a\in\mathbb{Z}$.
I've done it few, like 
$a=0\implies a^2+1=1$,
$a=\pm 1\implies 3-(a^2+1)=1$,
$a=\pm 2\implies 2\cdot 3-(a^2+1)=1$,
$a=\pm 3\implies (a^2+1)-3\cdot 3=1$,
$a=\pm 4\implies 6\cdot 3-(a^2+1)=1$....
At first, I thought it has some rule like $|a\cdot 3-(a^2+1)|=1$, but it broke at $a=\pm 4$.
So it does look like to have some combination of generators to make $1$ for each $a$, I don't know what's the general rule.
And I found that this evaluation function is a homomorphism : $\phi_a(fg)=f(a)g(a)=\phi_a(f)\phi_a(g)$, from $\mathbb{Z}[x]$ to $\mathbb{Z}$, so I should look for $\ker \phi_a\subset I$... If $\ker\phi_a\cap I\not=\emptyset$ for all $a$, then I'm done (since $1\not\in I$).
So I guess there're at least two ways... one is to find a general formula or rule for $f$, and the other is to show that the kernel of the evaluation function intersects $I$... and both are difficult for me now.
Any hint will be appreciated.
PS: Is my answer for 1) correct?
 A: I'm honestly not sure what you're trying to say in (1).
You're correct in that you want to show that $1\not\in I$, but I'm not sure what the whole thing with $1=h(x)\sum_k 3^{i_k}(x^2+1)^{j_k}$ is, or where it came from. The correct thing to say is that the elements of $I$ are of the form $3f+(x^2+1)g$, with $f,g\in\Bbb{Z}[x]$. Thus if $1\in I$, we have $1=3f+(x^2+1)g$. Reducing this mod 3, we see that $(x^2+1)g \equiv 1 \pmod{3}$, which is nonsense, since $\Bbb{Z}_3$ is a domain (indeed a field), so positive degree polynomials are not units in $\Bbb{Z}_3[x]$. Thus this is impossible.
You have the right idea for 2, but it can be simplified a bit. Note that $f_a(I)$ always contains 3, so to show that $f_a(I)=\Bbb{Z}$, it suffices to show that for any $a$, $f_a(I)$ contains an integer that is $\pm 1$ mod $3$. This is easier than trying to solve for $1$ (though I think that could also be done).
We have two cases:


*

*$a\equiv 0\pmod{3}$, then $a^2+1\equiv 1\pmod{3}$.

*$a\equiv \pm 1 \pmod{3}$, then $a^2+1\equiv 2\pmod{3}$.


We're done.
A: For the first question, compute the quotient ring. We have $\mathbb{Z}[x]/I \cong \mathbb{Z}/(3)[x]/(x^2+1)$ which is nonzero (why?). 
For the second problem, notice that $x^2+1$ evaluated at $1$ is $2$, so the set contains all linear combinations of two coprime integers. 
