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

1. myininaya

$\text{ let } k=\frac{n^2}{2n+1} \\ k(2n+1)=n^2 \\ n^2-2kn-k=0 \\ n=\frac{2k \pm \sqrt{4k^2+4k}}{2} \\ \text{ so } n \text{ is an integer } \\$ just showing my thinking and this thinking though

2. myininaya

$n=k \pm \sqrt{k^2+k} \\ n=k \pm \sqrt{k(k+1)} \\ k>0$ so I guess the equivalent question here is to show for integer k>0 we have k(k+1) is not a perfect square

3. myininaya

I 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

4. myininaya

example: 1*2 2*3 3*4 4*5 5*6 6*7 7*8 both factors would have to be the same

5. myininaya

of course k(k+1) is a perfect square when k=0 or k=-1

6. myininaya

but as you said k>0

7. beginnersmind

Seems to be correct, though not the proof I had in mind.

8. beginnersmind

I'll leave this up with the hint that there's a 3 line proof if someone else wants to try it.

9. thomas5267

$\frac{n^2}{2n+1}=\frac{(n+1)^2-2n-1}{2n+1}=\frac{(n+1)^2}{2n+1}-1$ $$n$$ and $$n+1$$ shares no common factors?

10. beginnersmind

I'm not sure I see how that helps.

11. beginnersmind

Ok, 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.

12. beginnersmind

Which 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

13. thomas5267

Pretty sure that $$n$$ and $$n+1$$ shares no common factor. $n=p_1^{k_1}p_2^{k_2}\cdots\\ n+1=p_1^{k_1}p_2^{k_2}\cdots+1\\ n+1=1\pmod {p_k}$

14. beginnersmind

Yeah, 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.

15. thomas5267

$$2n+1$$ 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 $$(n+1)^2-(2n+1)=(n+1)^2-0\not\equiv 0\pmod{2n+1}$$ 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.

16. beginnersmind

Right, 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 d|1 => d = 1.

Find more explanations on OpenStudy