How could I prove that $x^2+1$ and $x^4+1$ are irreductible over $\mathbb{Z}/2 \mathbb{Z}$ and $\mathbb{Z}/3 \mathbb{Z}$?

Question

How could I prove that $x^2+1$ and $x^4+1$ are irreductible over $\mathbb{Z}/2 \mathbb{Z}$ and $\mathbb{Z}/3 \mathbb{Z}$?

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) (\mod I)$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreductible in $A[x]$

I think I have to use this theorem, but I am not certain. Is anyone could help me at this point?

Answer 1

.$x^2+1=x^2+1^2=(x+1)^2$ in $2\mathbb{Z}$

$x^4+1 = (x^2+x+2)(x^2+2x+2)$ in $3\mathbb{Z}$

Neither is irreducible mod it's respective field. In fact, $x^4+1$ is reducible mod $n\mathbb{Z}$ for every $n$!