Question

Find a prime p > 5such that x²+ 1is reducible in${\mathbf{\mathbb{Z}}}_{\mathbf{p}}\left[x\right]$.

Short answer

It is proved that the required prime number is $p=7$.

Step-by-step solution

Factor Theorem:

If F is any field such that $\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left[x\right]$and $\mathit{k}\mathbf{\in }\mathit{F}$. Then, k will be the root of the polynomial f(x), if and only if role="math" localid="1648623679350" $(x-k)$is a factor of f(x)in F[X].

Irreducible Polynomial:

The given polynomial is $f\left(x\right)=x^2+1$

Let’s take a prime number, $p=7$.

Now, checking for the roots as:

$$\begin{gathered} f\left(0\right)=0^2+1=1\ne 0 \\ f\left(1\right)=1^2+1=2\ne 0 \\ f\left(2\right)=2^2+1=5\ne 0 \\ f\left(3\right)=3^2+1=10=3\ne 0 \\ f\left(4\right)=4^2+1=2+1=3\ne 0 \\ f\left(5\right)=5^2+1=4+1=5\ne 0 \\ f\left(6\right)=6^2+1=1+1=2\ne 0 \end{gathered}$$

Clearly, these are irreducible polynomials.

Hence, the required prime number is $p=7$.