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beginnersmind
 one year ago
For fun:
If n is a positive integer prove that the fraction \(\frac{n^2}{2n+1}\) cannot be simplified.
beginnersmind
 one year ago
For fun: If n is a positive integer prove that the fraction \(\frac{n^2}{2n+1}\) cannot be simplified.

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ let } k=\frac{n^2}{2n+1} \\ k(2n+1)=n^2 \\ n^22knk=0 \\ n=\frac{2k \pm \sqrt{4k^2+4k}}{2} \\ \text{ so } n \text{ is an integer } \\ \] just showing my thinking and this thinking though

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2\[n=k \pm \sqrt{k^2+k} \\ n=k \pm \sqrt{k(k+1)} \\ k>0 \] so I guess the equivalent question here is to show for integer k>0 we have k(k+1) is not a perfect square

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2I think it makes sense to say since to say there is no two consecutive positive integers such that when multiplied give us a perfect square

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2example: 1*2 2*3 3*4 4*5 5*6 6*7 7*8 both factors would have to be the same

myininaya
 one year ago
Best ResponseYou've already chosen the best response.2of course k(k+1) is a perfect square when k=0 or k=1

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Seems to be correct, though not the proof I had in mind.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0I'll leave this up with the hint that there's a 3 line proof if someone else wants to try it.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \frac{n^2}{2n+1}=\frac{(n+1)^22n1}{2n+1}=\frac{(n+1)^2}{2n+1}1 \] \(n\) and \(n+1\) shares no common factors?

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure I see how that helps.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I sort of see it now. If n+1 and n share no common factors that neither do (n+1)^2 and n^2. So n^2 can't have a common factor with 2n+1 because that common factor would be a common factor between n^2 and n^2+2n+1, which is (n+1)^2. It's actually similar to the solution I had.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Which is: gcd(2n+1,n) = 1 so n and (2n+1) share no prime factor. therefore n^2 and (2n+1) share no prime factor

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Pretty sure that \(n\) and \(n+1\) shares no common factor. \[ n=p_1^{k_1}p_2^{k_2}\cdots\\ n+1=p_1^{k_1}p_2^{k_2}\cdots+1\\ n+1=1\pmod {p_k} \]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, n and n+1 definitely share no common factor. :) I was trying to reconstruct how you used that fact to prove that n^2 and 2n+1 share no common factor.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\(2n+1\) cannot share factors with \(n\) and \(n+1\) at the same time so there is at most one of them is reducible. WLOG, let \(\dfrac{n^2}{2n+1}\) be reducible and \(\dfrac{(n+1)^2}{2n+1}\) be irreducible. So \(n^2\equiv 0\pmod{2n+1}\) and \((n+1)^2\not\equiv0\pmod{2n+1}\). But \((n+1)^2(2n+1)=(n+1)^20\not\equiv 0\pmod{2n+1}\) and \((n+1)^2(2n+1)=n^2\equiv 0\pmod{2n+1}\). So it must be the case that both \(\dfrac{n^2}{2n+1}\) and \(\dfrac{(n+1)^2}{2n+1}\) are irreducible.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0Right, got it. Another way to show that n and (2n+1) share no common factors, is by realizing that any common factor d divides the expression (2n+1)  2(n) = 1. So d1 => d = 1.
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