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Show that $X^5+X^3+1$ is irreducible over $\mathbb{Z}$ and $X^3+aX^2+bX+1$ is reducible iff $a=b$ or $a+b=-2$.

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What do you mean "irregularly polynomial"? Perhaps you meant "irreducible polynomial"? – ybungalobill Dec 25 '12 at 16:48
ok, sorry anyone – Muniain Dec 25 '12 at 17:02
I think you might have the conditions on $X^3+aX^2+bx+1$ backwards - this polynomial is reducible iff $\dots$ – Thomas Andrews Dec 25 '12 at 17:36

A cubic polynomial that is monic in ${\mathbf Z}[x]$ is reducible if and only if it has a root in the integers, and for your particular cubic the only possible roots in the integers could be $\pm 1$ (I'll let you figure out why that is). Set $X$ equal to $1$ and $-1$ in the polynomial to see what constraint is imposed by asking for these numbers to be roots.

As for the quintic, could you please tell us what irreducibility tests you know? The question sounds like homework, because it is written as a command "Show that..." with no background context whatsoever and it seems strange that you would want to know why that one quintic is irreducible and why that parametric family of cubics is irreducible at the same time. What have you tried yourself?

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So, for example, do you know any methods involving finite fields? – Mark Bennet Dec 25 '12 at 18:50
I can't apply Eisenstein's criterion because coefficient of $x^3$ is $1$. So my strategies is shift $x$. But power of $x$ is very big. How should I shift $x$? – Muniain Dec 26 '12 at 8:47
I am not sure, but replace $x$ with $x+1$ and check if you can use Eisenstein's criterion, another hint is use derivative test to locate roots if any.(I can't give more hints for first, and for second, another brute force method is, if Cubic is reducible, it has atleast one linear factor. – Ram Dec 26 '12 at 13:07
@Firmino: It is impossible to use the Eisenstein criterion: replacing $x$ with $x+c$ for an integer $c$ will never turn $x^5+x^3+1$ into an Eisenstein polynomial at a prime. What other irreducibility tests do you know besides the Eisenstein criterion? – KCd Dec 28 '12 at 3:33

Hint for the first part. You might notice something about the parity of $x^5+x^3+1$ which would imply something about the parity of factors - equivalent to working mod 2.

Suppose you have a factorisation in $\mathbb Z$, then you can reduce that factorisation mod $n$ for any $n \in \mathbb Z$ to get a factorisation mod $n$ (generally useful for primes $p$). So if you can find a $p$ for which there is no factorisation mod $p$, then there is no factorisation in $\mathbb Z$.

You can also use this to prove that there is no factorisation of a particular kind (eg cubic times quadratic).

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