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Proving Polynomial Irreducibility over Z

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Homework Statement


Prove that the polynomial $(x-1)(x-2)...(x-n) + 1$ is irreducible over Z for n\geq 1 and n \neq 4


Homework Equations


N/A


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!
 

Answers and Replies

  • #2
Dick
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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
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??
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
Dick
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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.
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
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
Dick
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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.
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?
 
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