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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- beginnersmind

- schrodinger

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

- 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

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- myininaya

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

- myininaya

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

- myininaya

but as you said k>0

- beginnersmind

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

- beginnersmind

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

- 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?

- beginnersmind

I'm not sure I see how that helps.

- 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.

- 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

- thomas5267

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

- 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.

- 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.

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.