# Thread: Reducible problem. Help!

1. ## Reducible problem. Help!

Suppose p>2 is a prime and $x^2+1$ is reducible in Fp[x]. Prove that the multiplicative group (Fp\{0}, $\cdot$) contains an element of order 4. Deduce that p $\equiv$1mod4.

I'm finding algebra quite hard atm, and this question has got me stuck. I don't know where to start. Please help!

2. Originally Posted by gizmo
Suppose p>2 is a prime and $x^2+1$ is reducible in Fp[x]. Prove that the multiplicative group (Fp\{0}, $\cdot$) contains an element of order 4. Deduce that p $\equiv$1mod4.

I'm finding algebra quite hard atm, and this question has got me stuck. I don't know where to start. Please help!

$x^2+1$ reducible mod p $\Longrightarrow x^2=-1\!\!\!\pmod p$ has a solution in $\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}$.

Now, $x^2=-1\!\!\!\pmod p\Longrightarrow x^4=1\!\!\!\pmod p\Longrightarrow x\in\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}$ has order 4 $\Longrightarrow 4\mid ord\left(\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}\right)|=p-1\Longrightarrow$ ...end the argument now.

Tonio

3. Thank you so much, that makes sense to me now! Hopefully I'l be able to make some progress through the rest of my algebra

4. Originally Posted by tonio
$\Longrightarrow 4\mid ord\left(\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}\right)|=p-1\Longrightarrow$ ...end the argument now.

Tonio
Sorry, I followed up to this point, but I'm a bit confused as to what this first term means?

5. Originally Posted by gizmo
Sorry, I followed up to this point, but I'm a bit confused as to what this first term means?

"Four divides the order of the multiplicative group of non-zero residues modulo p, whose order is p-1"

Tonio

6. thanks! i think i get this now, so to finish it off...
$4|p-1 \Rightarrow p-1 \equiv 0mod4 \Rightarrow p \equiv 1mod4$