My Math Forum Help needed - UFD

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 February 16th, 2011, 08:23 AM #1 Senior Member   Joined: Dec 2007 Posts: 687 Thanks: 47 Help needed - UFD Ok, so I came across this problem and I think it is too difficult, so I'm asking for you guys help! the problem is to find integer solutions to $n^3+n-1=x^2$ first of all three solutions are n =1, 2, 13, I don't know if there is any other, probably not lets factorize the whole thing like this: $n(n^2+1)=x^2+1=(x-\sqrt{-1})(x+\sqrt{-1})=n(n-\sqrt{-1})(n+\sqrt{-1})$ $\mathbb{Q}[\sqrt{-1}]$ is a UFD (http://en.wikipedia.org/wiki/Stark%E2%8 ... er_theorem), so we have $gcd(n,\ n^2+1)=gcd(x-\sqrt{-1},\ x+\sqrt{-1})=1$ from here I'm not sure how to proceed exactly, but I think it is clear that for n>1 one of $(x-\sqrt{-1})(x+\sqrt{-1})$ must be reductible, but it gives too many cases to check another possibility is to write $(\sqrt{n(n^2+1)}-1)(\sqrt{n(n^2+1)}+1)=x^2$ Do the same the same argument of UFD holds? If yes, I think $gcd(\sqrt{n(n^2+1)}-1,\ \sqrt{n(n^2+1)}+1)=1$, yielding: $\sqrt{n(n^2+1)}-1=a^2$ and $\sqrt{n(n^2+1)}+1=b^2$ where $a^2b^2=x^2$ I'm not sure how to proceed with the second factorization since $\sqrt{n(n^2+1)}$ is clearly irrational, do you guys see any mistake? thanks in advance!
February 16th, 2011, 11:32 AM   #2
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Re: Help needed - UFD

Quote:
 Originally Posted by al-mahed $n(n^2+1)=x^2+1=(x-\sqrt{-1})(x+\sqrt{-1})=n(n-\sqrt{-1})(n+\sqrt{-1})$ $\mathbb{Q}[\sqrt{-1}]$ is a UFD (http://en.wikipedia.org/wiki/Stark%E2%8 ... er_theorem), so we have $gcd(n,\ n^2+1)=gcd(x-\sqrt{-1},\ x+\sqrt{-1})=1$
I'm afraid that I don't understand this conclusion. Are you saying that if ab = cd in a UFD, then gcd(a,b) = gcd(c,d)? That's not true in the rational integers: (6)(15) = (5)(1, but gcd(6,15) = 3 and gcd(5,1 = 1.

February 16th, 2011, 11:45 AM   #3
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Re: Help needed - UFD

Quote:
 Originally Posted by al-mahed $gcd(n,\ n^2+1)=gcd(x-\sqrt{-1},\ x+\sqrt{-1})=1$
This isn't true in generality. $\gcd(n,n^2+1)=1$ in the Gaussian integers just like in the integers, but $\gcd(x-i,x+i)$ could be any divisor of 2 depending on the value of x.

LaTeX note: You don't need \$ inside the tags, they're in math mode already. Also, the code for gcd is \gcd, not gcd.

 February 16th, 2011, 02:50 PM #4 Senior Member   Joined: Dec 2007 Posts: 687 Thanks: 47 Re: Help needed - UFD Hmmmm, but by suposing there is a prime $\pi$ dividing both we get a contradiction, right? we get $\pi | 2x,\ \pi | 2\sqrt{-1}$... I see, there is one more condition, I need to take the Norms and see what divides what and if it is contradictory, I supose.
February 16th, 2011, 03:04 PM   #5
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Re: Help needed - UFD

Quote:
 Originally Posted by al-mahed Hmmmm, but by suposing there is a prime $\pi$ dividing both we get a contradiction, right?
No. If x = i, then gcd(i - i, i + i) = 2 (up to associates, of course; you could just as easily say 2i or -2). Can you find one that gives 1 - i? I can't immediately see one (though I haven't looked hard) nor can I immediately prove it does not occur.

February 16th, 2011, 03:31 PM   #6
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Re: Help needed - UFD

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Originally Posted by CRGreathouse
Quote:
 Originally Posted by al-mahed Hmmmm, but by suposing there is a prime $\pi$ dividing both we get a contradiction, right?
No. If x = i, then gcd(i - i, i + i) = 2 (up to associates, of course; you could just as easily say 2i or -2). Can you find one that gives 1 - i? I can't immediately see one (though I haven't looked hard) nor can I immediately prove it does not occur.
but we need to consider x integer

to the second question, you mean gcd=1-i? if yes $- i*( 1 + i )= - i + 1$

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