Z.G.

Does anyone know of any introductory books on elliptic functions, which can be read by someone who knows only calculus and are of a more historical flavour? By "historical flavour" I mean that the book should go according the historic development of the subject.

Good night

See you around

I see

It is indeed orders of magnitude better.

Yeah that's pretty good, let me see what happens with ordinary binet

@PM2Ring is there something special with those?

1000th I mean

Even the hundredth Fibonacci number is insane

It's remarkable that we still don't know whether the Fibonacci sequence contains infinitely many primes.

@PM2Ring Like what's the problem with good old Binet's formula?

I mean can you state the procedure?

What would that be like?

Hmm

@PM2Ring Guess I won't get any sleep today.

Hmm

I see

Oh

@Thorgott that the solutions differ by solutions to ax+by=0

@PM2Ring The case where GCD = 1 is the one I had in mind (I was proving that two consecutive Fibonacci numbers are coprime using Bezout's lemma, that's horrible but I managed to prove it simply by contradiction)

@Thorgott Why is that true?

My bad

@Thorgott Hmm I didn't read PM's reply properly and assumed that that's what he wrote.

@Thorgott wiki says it is

@PM2Ring Thanks anyways.

@PM2Ring Also, what keywords did you use to find that?

Does it have something to do with the fact that $ax+by=c$ is a line?

@PM2Ring thanks! I didn't know the addition law, you have saved me hours of point less labour.

How do you know it gives all?

I've seen F_m|F_n iff(?) m|n, where F_k is the k-th Fibonacci number, multiple times but never with proof, any hints?

What I'm trying to say is that: can every solution of ax + by = k be "reached" by the one obtained by Euclid's algorithm.

Is that a stupid question again?

By saying something I mean is there no relation between x, y and x', y'?

So if I find a and b using Euclid's algorithm for the first equation and I am given the equation ax' + by' = k can I say nothing about x' and y'?

I'm drunk I guess

Now if k is the GCD of a and b?

Oh nevermind

Then what values satisfy the second equation?

@SoumikMukherjee Why not?

k, a and b are positive.

Same a, b and k.

Everything is an integer in the second equation too.

If we are given ax + by = 1 (everything is an integer), then obviously a(kx) +b(ky) = k, now if we are given another equation ax' + by' = k, then is it necessary that x' and y' are multiples of k?

@PM2Ring Interesting I'll try to prove that.

I'm trying to prove that a^(1/m) + b^(1/n), where a, b, m and n are natural, and (a, b) and (m, n) are coprime pairs, is always irrational, can anyone give a small hint?

@SohamSaha Good Night

The entrance route is fine, how did INMO go btw?

@SohamSaha haha, I don't like sitting for exams unless absolutely necessary.

But I don't think I'll sit for that, I mean I suck at chemistry.

@user21820 Thanks, seems like I'll have to do a lot of work. Also I wasn't expecting you to give sources for calculus, I think I've got some good books and it's not as hard as the oly subjects involved.