Question

Prove that x² + 1 is reducible in ${\mathbf{\mathbb{Z}}}_{\mathbf{p}}\left[x\right]$if and only if there exist integers a and b such that $\mathit{p}\mathbf{}\mathbf{=}\mathbf{}\mathit{a}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{a}\mathit{b}\mathbf{=}\mathbf{1}\mathbf{}\left(modp\right)$.

Short answer

Hence, the given statement is proved.

Step-by-step solution

Determine whether integers a and b exists so that, a+b=0 (mod p) and $a + b = p$     

Let’s consider that, x² + 1 is reducible in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

By using theorem 4.11, x² + 1 an be written as a product of two polynomials in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$ of lower degrees. So that,

$x^2+1=\left(ax+c\right)\left(bx+d\right)$

From this, it is concluded that, cd = 1.

Remember that without any loss of generality one can suppose that c = 1 in and d = 1 in ${\mathrm{\mathbb{Z}}}_p\left[x\right]$.

Now, multiply equality ( * ) by $c^{-1}and{d}^{-1}$ from the left hand side. Then,

$$\begin{gathered} c^{-1}{d}^{-1}\left(x^2+1\right)=c^{-1}\left(ax+c\right)d^{-1}\left(bx+d\right) \\ =\left(c^{-1}ax+1\right)\left(d^{-1}bx+1\right) \\ =x^2+1 \end{gathered}$$

Therefore,

$x^2+1=\left(ax+1\right)\left(bx+1\right)=\left(ab\right)x^2+\left(a+b\right)x+1$

After that, compare the coefficients in polynomials on both hand side, then $ab=1\left(\mathrm{mod}p\right)$and that $a+b=0\left(\mathrm{mod}p\right)$.

As $0<a,b<p$ it follows that $0<a+b<2p$. Therefore, it would be a + b = p.

Thus it is proved that, there exists integers a and b so that, $a+b=0\left(\mathrm{mod}p\right)$and a + b = p.

Determine x2 + 1 is reducible in  ℤp[x]

There exists integers a and b so that, $a+b=0\left(\mathrm{mod}p\right)$and a + b = p.

Note that,

$\left(ax+1\right)\left(bx+1\right)=\left(ab\right)x^2+\left(a+b\right)x+1=x^2+1in{\mathrm{\mathbb{Z}}}_p$

As either a or b ca be$0in{\mathrm{\mathbb{Z}}}_p$.

Therefore,x² + 1 is reducible in Z_p.

Hence proved.