# Homework Help: Proving Polynomial Irreducibility over Z

1. Nov 17, 2011

### Discover85

1. The problem statement, all variables and given/known data
Prove that the polynomial $(x-1)(x-2)...(x-n) + 1$ is irreducible over Z for n\geq 1 and n \neq 4

2. Relevant equations
N/A

3. The attempt at a solution
Let $f(x) = (x-1)(x-2) \cdots (x-n) + 1$ and suppose $f(x) = h(x)g(x)$ for some $h,g \in \mathbb{Z}[x]$ where $\deg(h), \deg(g) < n$. Note that $f(x) = 1$ for $x \in (1,2,\ldots,n)$ and so $h(x)g(x) = 1$ for $x \in (1,2,\ldots,n)$. This implies that $h(x) = \pm 1$ and $g(x) = \pm 1$ for those $x$ values. Moreover, we must have $g(x) - h(x) = 0$. Since a polynomial of degree $m$ is determined by $m+1$ points we have that $h(x) = g(x)~\forall x$. This implies that $f(x) = g(x)^2$. Now consider $f(n+1) = g(n+1)^2$. We have that $n! + 1 = j^2$ where $j \in \mathbb{Z}$. I feel like I'm close but don't know where to go from here... Any help will be greatly appreciated!

2. Nov 17, 2011

### Dick

That's really very clever. You've reduced it to Brocard's problem. The trouble is that you've reduced it to a really hard problem. The known solutions are that 4!+1, 5!+1 and 7!+1 are perfect squares. But whether there are more is still, as far as I know, an open question. BTW you don't have to worry about 5 or 7. n!+1=j^2 isn't quite the right equation. You really only have to worry about n even. Maybe that has an easier solution??

3. Nov 17, 2011

### Discover85

Yeah, I realised I had reduced it to something fairly ugly when I was using Wolfram and found 7! + 1 fit the bill. I don't know if this helps but this question was posed under the Eisenstiens Criterion section.

4. Nov 17, 2011

### Dick

Well, 7!+1 is a perfect square but that's still not an exception to irreducibility since you would need -7!+1 to be a perfect square. And it's not. You've got a sign missing in the n!+1=j^2 criterion. Odd values of n are really no problem. It's the even ones I'm worried about. And I've been scratching my head over this for a while and I don't know what to do with it. A trick you can sometimes pull with the Eisenstein Criterion is that p(x) is irreducible iff p(x+k) is irreducible for every integer k. But I'm not getting that to go anywhere either. Maybe it's just late and I'm thinking the wrong way about it. You try again and we'll compare notes.

5. Nov 17, 2011

### Discover85

From my proof we know that for n<4 the polygon is irreducible since n! + 1 != m^2. As for n>5, you can take the derivative and show that the polygon has a minimum less than 0. This contradicts the fact that f(x) = g(x)^2. Thus n = 4.

6. Nov 17, 2011

### Dick

Ok, the minimum thing works. That's also really clever! I can see how that works by substituting a value like x=n-1/2. How does using derivatives prove it more easily? Doesn't have much to do with Eisenstein's criterion, does it?

Last edited: Nov 18, 2011