Difference Between Divisors

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Abdullah Al Tanzim
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Difference Between Divisors

Unread post by Abdullah Al Tanzim » Wed May 15, 2019 2:12 pm

Show that there are infinitely many positive integers $n$ such that $n^2+1$ has two positive integer divisors whose difference is $n$.
Last edited by Abdullah Al Tanzim on Wed May 15, 2019 2:39 pm, edited 1 time in total.
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein

User avatar
Abdullah Al Tanzim
Posts: 20
Joined: Tue Apr 11, 2017 12:03 am
Location: Dhaka, Bangladesh.

Re: Difference Between Divisors

Unread post by Abdullah Al Tanzim » Wed May 15, 2019 2:34 pm

Solution:
We will prove that there are infinitely many positive integers $n$ and $a$ such that $n^2+1=a(a-n)$.

LEMMA: There are infinitely many positive integer solutions to the equation $5x^2-4=y^2$.
Proof: Assume that $(x',y')$ is a solution of this equation.Then we follow the transformation:
$(x,y) \mapsto (\frac{3x'+y'}{ 2}),(\frac{5x'+3y'}{2})$.
It's easy to check that this works.
AS $(1,1)$ is a solution of this equation, we can say there are infinitely many solutions of this equation.


Now, $n^2+1=a(a-n)$
$ \Rightarrow n^2+an+(1-a^2)=0$
$\Rightarrow 2n=\sqrt{5a^2-4}-a$ (which is due to the quadratic formula)
As we can see, we are left to prove that there are infinitely many positive integers $a$ such that $5a^2-4$ is a square number which is evident from our lemma.
So, we are done. ;) ;)
Everybody is a genius.... But if you judge a fish by its ability to climb a tree, it will spend its whole life believing that it is stupid - Albert Einstein

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