Question

I've searched a lot about these Elliptic curves...but i cant find a suffice answer :/

Idea of this question comes from workin on this problem :

What kind of number will result from \(\sqrt{n+\sqrt[3]{n+1}}\) where \( n \in \mathbb{N}\).

Answer 1

there are some positive integers that makes it integer...

\(n=7 \Rightarrow \sqrt{7+\sqrt[3]{7+1}}=2 \)

\(n=2196 \Rightarrow \sqrt{2196+\sqrt[3]{2196+1}}=47 \)

this got me thinking on how to find all positive integers that makes

\(\sqrt{n+\sqrt[3]{n+1}}\)
an integer.
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let
\(m=\sqrt{n+\sqrt[3]{n+1}}\)

First of all... in order for \(m\) to be an integer, also \(n+\sqrt[3]{n+1}\) has to be integer, and therefore \(\sqrt[3]{n+1}\) has to be integer. Let's call this number \(a\).

We have (because \(a=\sqrt[3]{n+1}\) ) :
\(n=a^3-1\)

And substituting in the original equation we get:

\(m^2=a^3+a-1\)

Now solving the original equation is equivalent to solve this Diophantine (considering that \(m\) is positive...) .

But there is a problem: the equation found is an elliptic curve...

Answer 2

this means that we have to find all the points having integer coordinates on this elliptic curve.(o.O) I'm sorry, I can't go further...

Answer 3

@KingGeorge
@Neemo
@eliassaab

Answer 4

what math are you in

Answer 5

this is something related to diophantine equations

Answer 6

saura that problem was a result of workin on this

Answer 7

Oh... ya

Answer 8

I can't help you with the algebra (I hate number theory more than nearly anything else), but if you just want a lot of pairs to play around with I can very quickly write a program to find them...

Answer 9

:) thank u very much...actually i want to know if there is an analyticall approach or not.

Answer 10

I've been reading some of the literature on this...(a) there are only finitely many solutions.

Answer 11

There is an algorithm by Alan Baker to find all of them.

Answer 12

so there are finite solutions for that...is that for all elliptic curves like this :\[y^2=x^3+Ax+B\]?

Answer 13

Yep. Those are called Weierstrass functions (don't confuse this with his everywhere continous but nowhere differentiable function).

Answer 14

I'm going to assume you know at least a little group theory. I don't know about finding integer solutions, but the rational solutions (as points) form a group, with composition of two points defined as the third point of intersection of the line they form and the original function. You can even find the tangent line at a single point, and where it intersects the graph again will be a rational solution (essentially P+P). I don't quite understand why this works...but it does. In any event, this gives you a (painful) way to get solutions, and you could in theory find the entire group (which you could then prove is the entire group by finding every composition).

Answer 15

It's also finitely generated...so you could find the generating set (no idea how to, though).

Answer 16

This is all coming from a paper someone wrote (see attached).

Answer 17

Thank you so much for all your help I really appreciate it :)

Answer 18

now i must say i really want to go into group theory

Answer 19

It's a wonderful subject. I find it much more fun than analysis or number theory. You could do worse than to get Algebra by Michael Artin...it's a very good textbook on Linear Algebra and group theory.

Answer 20

Nice approach @mukushla

Answer 21

oh ... thank u...but the main part is still a mystery

Answer 22

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