Prove that certain elements are not in some ideal I have the following question:

Is there a simple way to prove that

$x+1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$ and

$x-1 \notin \langle2, x^2+1\rangle_{\mathbb{Z}[x]}$

without using the fact that $\mathbb{Z}[x]/\langle2, x^2+1\rangle$ is an integral domain?

Thanks for the help.
First a comment: it looks like the goal of the problem is to show that $(2,x^2+1)$ isn't a prime ideal (and hence the quotient isn't a domain), because $(x+1)^2=x^2+1+2x$ is in the ideal, and we hope to show that $x+1$ isn't in the ideal.

If $x+1\in (2,x^2+1)$, then we would be able to find two polynomimals in $\Bbb Z[x]$, say $a$ and $b$, such that $x+1=2a+(x^2+1)b$. 

Looking at the equation mod 2, you get $x+1=(x^2+1)\overline{b}$, where all the terms are in $\Bbb F_2[x]$. But since $\Bbb F_2[x]$ is a domain, the degrees on both sides have to match. If $\deg(b)&gt;0$, then the degree of the right hand side would be at least 3, and even if the degree of $b$ were 0, the right hand side would have degree 2. It is impossible then, for such an expression to be equal to $x+1$. By this contradiction, we conclude $x+1$ is not in the ideal.

Finally, you can note that $x+1$ is in the ideal iff $x-1$ is, since $x-1+2=x+1$.

Thus we have shown that $(2,x^2+1)$ is not a prime ideal, and it isn't even a semiprime ideal.
$x-1 \in I \implies x-1=2p+q(x^2+1) \implies q=0$ (reasoning on degree) so that you have $x-1=2p$ which is clearly impossible.