Question

For fun:

If n is a positive integer prove that the fraction \(\frac{n^2}{2n+1}\) cannot be simplified.

Answer 1

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

Answer 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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

but as you said k>0

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

I'm not sure I see how that helps.

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

\(2n+1\) 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.

Answer 16

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.