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Question. Prove that $f(x)=x^{5}-x+a$ is irreducible in $\Bbb Z[x]$ if $5\nmid a$.

My approach. If we let $f(x)=(x+a_0)(x^4+b_3x^3+b_2x^2+b_1x+b_0)$, we know that $a=a^5-a$. So $5 \mid a$ and it's contradictory to condition.

But can we let $f(x)=(x^2+a_1x+a_0)(x^3+b_2x^2+b_1x+b_0)$? I can't prove we can't. I've used Eisenstein's criterion, but it was helpless. I couldn't use it anywhere. Can somebody help me? (I don't want a generalized proof.)

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    $\begingroup$ The polynomial in the title is different from the polynomial in the first line of your question $\endgroup$ – Zubin Mukerjee Feb 5 '19 at 3:53
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    $\begingroup$ Sorry. It's a typo. $\endgroup$ – coding1101 Feb 5 '19 at 3:54
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Here's an approach: we show that it's irreducible over $\mathbb{Z}/5$, and that implies it's irreducible over $\mathbb{Z}$.

As a function on the integers mod $5$, $f$ is simply the constant function $a$. Of course, since we can interpolate a polynomial of degree $\le 4$ to match any function on those five points, that isn't going to solve our problems on our own. We thus consider possible factorizations.

First, writing $f$ as the product of a degree-1 polynomial and a degree-4 polynomial. Since every polynomial of degree $1$ has a root, this is impossible; the product would then have a root, and $f$ doesn't.

Second, writing $f$ as a product of a degree-2 polynomial $g$ and a degree-3 polynomial $h$. Multiplying and dividing by constants, we can assume WLOG that $g$ has leading coefficient $a$. Then, completing the square, we write $g(x)\equiv a\left((x-b)^2+c\right)$.
As before, $g$ must not have any roots. That forces either $c\equiv 2$ or $c\equiv 3$.

Case 1: $c\equiv 2$. We have $g(x)\equiv \begin{cases}2a&x-b\equiv 0\\3a&x-b\equiv\pm 1\\a&x-b\equiv \pm 2\end{cases}$. Dividing, $h(x)\equiv\begin{cases}3&x-b\equiv 0\\2&x-b\equiv\pm 1\\1&x-b\equiv \pm 2\end{cases}$.
What's an interpolating polynomial that matches those values? To get $3$ at $y=0$ and $2$ at $y=\pm 1$, we can simply take $3-y^2$. That would give us a value of $3-4\equiv -1$ at $\pm 2$, so we need to add $2$ there to make it work. For something that's zero at $y=-1,0,1$, we need a factor of $y^3-y$ - but then $2^3-2\equiv 1$ and $3^3-3\equiv -1$. Those values don't match; we'll have to multiply by $y$ again to get an even function with equal values at $\pm 2$. Specifically, $y^4-y^2\equiv 2$ at $\pm 2$. Combine the parts, and
$3-y^2+(y^4-y^2) \equiv \begin{cases}3&y\equiv 0\\2&y\equiv \pm 1\\1&y\equiv \pm 2\end{cases}$. Thus $h(x)\equiv 3-2(x-b)^2+(x-b)^4$ as functions on $\mathbb{Z}/5$.
But polynomial interpolation is unique; there's only one polynomial (up to equivalence mod 5) of degree $\le 4$ that matches a complete set of values. We have $h$ equivalent to a fourth-degree polynomial, so it can't have degree $3$, and there's no factorization.

Case 2: $c\equiv 3$. We have $g(x)\equiv \begin{cases}3a&x-b\equiv 0\\4a&x-b\equiv\pm 1\\2a&x-b\equiv \pm 2\end{cases}$. Dividing, $h(x)\equiv\begin{cases}2&x-b\equiv 0\\4&x-b\equiv\pm 1\\3&x-b\equiv \pm 2\end{cases}$.
The interpolating polynomial this time is $2+2y^2-(y^4-y^2)$, so $h(x)\equiv 2+3(x-b)^2-(x-b)^4$ as a function. By the same argument, that equivalence is also true as a polynomial, and there's no factorization.

With all cases covered, we're done - $f$ is irreducible over $\mathbb{Z}/5$, and therefore also irreducible over $\mathbb{Z}$.

[Edit - just saw that comment go up. This is a grubby special case of the linked material, without the deeper insights]

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  • $\begingroup$ Is it enough to write "$f(x)$ doesn't have any roots" without any explanation in the contests? $\endgroup$ – coding1101 Feb 5 '19 at 5:15
  • $\begingroup$ The explanation is the first sentence of the previous paragraph. Also, you already had the case of a potential linear factor down. I didn't see the need to elaborate. $\endgroup$ – jmerry Feb 5 '19 at 5:21
  • $\begingroup$ Oh, sorry. I didn't see that sentence. $\endgroup$ – coding1101 Feb 5 '19 at 5:22
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If $5\nmid a$, then $x^5-x+a$ is an irreducible polynomial over $\Bbb F_5$. Then it is irreducible over $\Bbb Z$, and then by Gauss's lemma, over $\Bbb Q$.

Proving that it is irreducible over $\Bbb F_5$ is essentially Artin-Schreier theory. If $\xi$ is a zero of $x^5-x+a$ in an extension field $k$ of $\Bbb F_5$ then $\xi^5=\xi-a$. But $\xi^5=\varphi(\xi)$ where $\varphi$ is the Frobenius automorphism. Then $\varphi^k(\xi)=\xi-ka$, and $\varphi^k(\xi)=\xi$ iff $5\mid k$. Therefore $\xi$ defines a degree $5$ extension of $\Bbb F_5$, and so $x^5-x+a$ is irreducible over $\Bbb F_5$.

One can replace $5$ here by any prime.

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