Ideal of $\mathbb Z[x] $ generated by $\mathit x^2+1$ and $ 3x+1$

Question

Let $\mathit I $$\,$ be the ideal of $\mathbb Z[x] $ generated by $\mathit x^2+1$ and $ 3x+1$. Find the positive integer $\mathit n$ such that $\mathit I$ $\cap \mathbb Z$ = $\mathit n$$\mathbb Z$

I find $(\mathit x^2+1)(9)+ (3x+1)(-3x+1)= 10 $, So It is expected that 'n =10' But I can't find for n=1, 2, 5

Any hint will be appreciated.

Answer 1

Hint: $\ (x^2\!+\!1)f + (1\!+\!3x)g = n\!\iff\! \bmod x^2\!+\!1\!:\ 1\!+\!3x\mid n.\,$ But $\,\Bbb Z[x]/(x^2\!+\!1)\cong \Bbb Z[i]\,$ so

$$ 1\!+\!3i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \underbrace{\dfrac{n}{1\!+\!3i}\in \Bbb Z[i]\iff \dfrac{n(1\!-\!3i)}{10}\in\Bbb Z[i]}_{\large \text{rationalize denominator}}\iff \color{#c00}{10\mid n}\ \ {\rm in}\ \ \Bbb Z\qquad$$