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Given is the polynomial $\varphi(X)=X^4+1$.

Now there are two tasks:

(1) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with $p \equiv1$ (mod 4).

(2) Show, that $ \varphi(X)$ is reducible in $\mathbb F_p [X]$, where $p$ is prime number with $p \equiv3$ (mod 4).

My ideas until now:

(1) I try to find $a,b \in \mathbb F_p [X]$, such that $X^4+1 = (X^2+a)(X^2+b)$. Not to use the middleterm $+cX$ was a hint. Now $(X^2+a)(X^2+b)=X^4+aX^2+bX^2+ab=X^4+(a+b)X^2+ab$. This should equal $X^4+1$, so I know, that $a+b \equiv 0$ mod p and $ab \equiv 1$ mod p. This means, that $a$ and $b$ must be units in $\mathbb F_p$, more exactly, $a$ is the multiplicative inverse element of $b$ in $\mathbb F_p$. $a+b \equiv 0$ reasons, that $a$ is the additive inverse to $b$.

But know I don't know, how I can go on with conclusion. I hope, somebody can help me.

user26857
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Myrkuls JayKay
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  • I nowadays don't normally instaclose questions as dupes when I have answered the duplicate target. This time I made an exception because the answer is CW. Do comment here, if you disagree - this "principle" is still developing :-/ +1 for showing your work. The dupe target is a bit difficult to find because the default search engine cannot handle TeX. – Jyrki Lahtonen Jun 24 '17 at 06:32

2 Answers2

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The two congruences give you $$b^2 \equiv -1 \pmod{p}$$ This is the same as asking when is $-1$ a quadratic residues mod $p$? If you are familiar with Legendre symbol, then perhaps you might have seen $$\left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}=\begin{cases}1 & \text{ if } p \equiv 1 \pmod{4}\\-1 & \text{ if } p \equiv 3 \pmod{4}\end{cases}$$

So if $p \equiv 1 \pmod{4}$, then we have $a,b$ such that $a+b \equiv 0$ and $ab \equiv 1$. The existence of such $a,b$ implies that $a^2 \equiv -1 \pmod{p}$. Therefore, $$x^4+1 =(x^2+a)(x^2-a)$$

part(2)

Claim: When $p \equiv 3 \pmod{4}$, then one of $2$ or $-2$ will be a quadratic residue. For this consider $2^{\frac{p+1}{4}}$. It can be shown that this will be the square root of either $2$ or $-2$.

If $2$ happens to be a quadratic residue mod $p$, i.e. if there exists an $a$ such that $a^2 \equiv 2 \pmod{p}$, then we can factor as follows: $$x^4+1 = x^4+1+a^2x^2-a^2x^2=(x^2+1)^2-a^2x^2=(x^2+1+ax)(x^2+1-ax).$$

Now consider the case when $2$ is not a quadratic residue, then try to see if you can prove that $-2$ will definitely be a quadratic residue. Then you can use the idea above to complete the factorization.

Anurag A
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  • No, I haven't seen this notation yet, but thank you for the hint. How can I use this for the task? – Myrkuls JayKay Jun 23 '17 at 21:11
  • @MyrkulsJayKay hint: so if $p\equiv 1 \mod{4}$, then such $a,b$ will exist so you can factor the polynomial. – Anurag A Jun 23 '17 at 21:13
  • Ok, thank you very much. Now the first part is solved. In the second part I've got to show, that it is reducible, but the solution in the Legendre notation is -1 if $p \equiv 3$, this would say, that the polynomial $\varphi$ is irreducible... But only, if I choose the decomposition, where I've got no middleterms with $cX$ and $dX$. So I will do the decomposition with longform $X^4+1=(X^2+aX+b)(X^2+cX+d)$ and hope, that this will give me a solution to construct $a,b,c,d$. – Myrkuls JayKay Jun 23 '17 at 21:24
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    @MyrkulsJayKay I have added the solution for part (2). See the main solution. – Anurag A Jun 23 '17 at 21:27
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Alternatively, $x^{p^2}-x=x\left(x^{p^2-1}-1\right)$ is divisible by $x^8-1=\left(x^4-1\right)\left(x^4+1\right)$, as $8\mid p^2-1$ for all odd primes $p$. As the field $\mathbb{F}_{p^2}$ with $p^2$ elements is the splitting field of $x^{p^2}-x$ and $\left[\mathbb{F}_{p^2}:\mathbb{F}_p\right]=2$, it follows that $x^4+1$ must factor into irreducible polynomials in $\mathbb{F}_p[x]$ of degree at most $2$.

Batominovski
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