Can a Fibonacci-like Series Prove Infinite Solutions for Diophantine Equation?

[ramsey2879]

Problem: Use Fibonacci like series (but not the same starting numbers) to prove that there are an infinite number of solutions to 5a^2 + 5ab + b^2 = 11^8 with a and b coprime. In I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.

 

[Gib Z]

$$T_n=T_{n-1} + T_{n-2}$$
Factorise.
$$(\sqrt{5}a +\sqrt{5}b)^2 - 4b^2 = 11^8$$. Perhaps that reminds you of something?

 

[Gib Z]

O, quote: I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.

Well, 11^8 = 123456787654321, which satisfies your conditions.

 

[ramsey2879]

O, quote: I have a proof that you can solve for any prime, ending in 1 or 9, multiplied to the 8th power.

Well, 11^8 = 123456787654321, which satisfies your conditions.

Actually the problem I meant is to find my proof that there are an infinite number of solutions to the Diophantine equation 5a^2 + 5ab + b^2 = p^n where a,b are coprime and p is a prime ending in 1 or 9. There is a relationship between any three consecutive term of a Fibonacci type series and the form 5a^2 + 5ab + b^2 that is invariant with the index number of the first term. That is a key to my proof. I leave the proof for you to figure out, but will make suggestions if you reach a dead end and have no idea where to turn.

First off then, how does the form 5a^2 + 5ab + b^2 relate to three terms of a Fibonacci type series in an invariant manner? (By Fibonacci type, I mean F(n) = F(n-1) + F(n-2)

 

[ramsey2879]

$$T_n=T_{n-1} + T_{n-2}$$
Factorise.
$$(\sqrt{5}a +\sqrt{5}b)^2 - 4b^2 = 11^8$$. Perhaps that reminds you of something?

This is interesting, it factors to $$((\sqrt{5}a +\sqrt{5}b +2b) * ((\sqrt{5}a +\sqrt{5}b -2b) = 11^8$$. However, these are not integers and I thought number theory dealt only with integers.

 

[matt grime]

That's a strange thing to say. Not least because 'a' isn't an integer, it is an indeterminate expression, so your own first post fails your own criterion.

Moving on from the stupidly pedantic, you're objecting to the use of sqrt(5), right? Well, if you think that sqrt(5) has no place in number theory then you're not going to get very far in the subject. Number theory might be set up to answer questions posed about Z, or more likely Q, but there is nothing that says it must stick only with those things, and indeed it doesn't.

 

[ramsey2879]

That's a strange thing to say. Not least because 'a' isn't an integer, it is an indeterminate expression, so your own first post fails your own criterion.

Moving on from the stupidly pedantic, you're objecting to the use of sqrt(5), right? Well, if you think that sqrt(5) has no place in number theory then you're not going to get very far in the subject. Number theory might be set up to answer questions posed about Z, or more likely Q, but there is nothing that says it must stick only with those things, and indeed it doesn't.

Ok but how would you get an solution in integers to my diophantine equation using that? If there is a way I would like to know. But since thare are Fibonacci identities that are known, it is not necessary to resort to the square root of 5 or the golden ratio to make my proof.

 

[matt grime]

I did not say anything about your equation. However, I believe it is a common tactic in algebraic number theory that in order to find integral solutions to something one passes to a larger ring, such as the ring of integers in a number field. This is probably called something like Class Field Theory.

 

[ramsey2879]

That's a strange thing to say. Not least because 'a' isn't an integer, it is an indeterminate expression, so your own first post fails your own criterion.

But isn't it understood that by talking about Fibonacci series and saying that a is coprime to b that we are talking about integers?

 

[matt grime]

Nope. coprimality exists in the algebraic integers too: x and y a coprime if z divides x and z divides y implies z is a unit. Anyway, you were asking for integral solutions to some equation in a and b. At least that is how I read it. But a was an indeterminate. Not an integer. 2 is an integer. a isn't. Unless a=2. For example. If a is an integer in post 1 what integer is it?

 

[matt grime]

http://www.maths.bris.ac.uk/study/undergrad/current_units/l3_units/algnumth.php

the general description tells you how it is sometimes advisable (indeed necessary) to move beyond the integers to find integer solutions, precisely by adding in square roots of numbers.

 

[ramsey2879]

I did not say anything about your equation. However, I believe it is a common tactic in algebraic number theory that in order to find integral solutions to something one passes to a larger ring, such as the ring of integers in a number field. This is probably called something like Class Field Theory.

But how do you get 5^a^2 + 5ab + b^2 from 5(a+b)^2 - 4b^2

 

[matt grime]

Uh? Please point out where you think I claim to have got that at all.

 

[ramsey2879]

Nope. coprimality exists in the algebraic integers too: x and y a coprime if z divides x and z divides y implies z is a unit. Anyway, you were asking for integral solutions to some equation in a and b. At least that is how I read it. But a was an indeterminate. Not an integer. 2 is an integer. a isn't. Unless a=2. For example. If a is an integer in post 1 what integer is it?

l a = 3161 b = 7159 , is one of infinitely many solutions in integers for 11 ^8. But any prime ending in 1 or 9 to any integer power can replace 11^8 and there would still be an infinite number of solutions in integers.

 

[ramsey2879]

Uh? Please point out where you think I claim to have got that at all.

You didn't post it but there was a post that used sqrt(5) in a manner that gets 5a^2 + 10ab - b^2 which is not my equation. Sorry

I have to go now. Thanks for your help.

 

[ramsey2879]

I did not say anything about your equation. However, I believe it is a common tactic in algebraic number theory that in order to find integral solutions to something one passes to a larger ring, such as the ring of integers in a number field. This is probably called something like Class Field Theory.

Are all primes ending in 1 or 9 and their products members of a ring? My proof related to this topic is much broader than the original proposition since it covers all primes ending in 1 or 9 or their products. In short I have a proof that there are an infinite number of solutions to the Diophantine equation
$$P = 5a^{2} + 5ab + b^{2}$$ where P is a prime ending in 1 or 9 or a product of such primes. Knowing the algorithm makes it easy to solve for any such product given initial solutions for the primes ending in 1 or 9. There is no need to do calculations greater than the square root of P. The proof is really simple in retrospect and can be understood by an algebra student familiar with the law of quadratic reciprocity. I doubt that it could be simplified by using Class Field Theory. Perhaps it may be a trival application of class theory but I don't know. I skipped college after taking the equivalent of two years of courses via night school (none in higher algebra or number theory) and am now retired. However, my interest in number theory is such that I would be willing to pay to audit a few courses such as Algebraic Number Theory once I take care of a few other things like building my dream home. The book I purchased is mostly Greek to me.

 

[matt grime]

Are all primes ending in 1 or 9 and their products members of a ring?

OF course they are: the integers, the rationals, the reals, complex numbers, algebraic integers, algebraic numbers...

 

[Gib Z]

perhaps the most obvious one, the primes >.<"

EDIT: btw, I am taking ring as a synonym for set, i haven't done rings yet but it looks like you were listing the sets primes were in.

 

[matt grime]

A ring is not just a set.

 

[Gib Z]

ahh then excuse my ignorance :)

 

[ramsey2879]

OF course they are: the integers, the rationals, the reals, complex numbers, algebraic integers, algebraic numbers...

I see that you didn't list sets such as all primes, etc as a ring.
Is a ring defined such that if "a" and "b" are members, then "a+b" and "a-b" are also members?
What is a unit in a ring?

 

[matt grime]

Google it. You'll get all the information you want and you won't have to keep posting questions.

But to answer the ones you have asked:

a ring is a set R, with two operations + and *, such that * distributes over +, (R,+) is an abelian group with identity 0. * is usually required to have an identity as well, and that is denoted by 1. * is an (associative) binary operation on R.

Think of + as addition, and * as multiplication.

Examples as above. Plus M_n the space of nxn matrices over some gadget (like another ring).

A unit is an element with an inverse under *.

In Z the units are just +/-1, in Q, C, R all nonzero elements are units. In M_n(Reals) the units are the invertible matrices (the ones with non-zero determinant).

 

[complexPHILOSOPHY]

perhaps the most obvious one, the primes >.<"

EDIT: btw, I am taking ring as a synonym for set, i haven't done rings yet but it looks like you were listing the sets primes were in.

Have you done any basic stuff with sets and functions, such as proving trivial, injection, surjection and bijection?

 

[ramsey2879]

Google it. You'll get all the information you want and you won't have to keep posting questions.

But to answer the ones you have asked:

a ring is a set R, with two operations + and *, such that * distributes over +, (R,+) is an abelian group with identity 0. * is usually required to have an identity as well, and that is denoted by 1. * is an (associative) binary operation on R.

Think of + as addition, and * as multiplication.

Examples as above. Plus M_n the space of nxn matrices over some gadget (like another ring).

A unit is an element with an inverse under *.

In Z the units are just +/-1, in Q, C, R all nonzero elements are units. In M_n(Reals) the units are the invertible matrices (the ones with non-zero determinant).

Then all integers from 0 to n-1 constitute a ring where multiplication and addition are taken mod n.

 

[Gib Z]

Ahh btw No I haven't done anythings about those stuff u listed complexPHILOSOPHY...

 

[ramsey2879]

Are all primes ending in 1 or 9 and their products members of a ring? My proof related to this topic is much broader than the original proposition since it covers all primes ending in 1 or 9 or their products. In short I have a proof that there are an infinite number of solutions to the Diophantine equation
$$P = 5a^{2} + 5ab + b^{2}$$ where P is a prime ending in 1 or 9 or a product of such primes. Knowing the algorithm makes it easy to solve for any such product given initial solutions for the primes ending in 1 or 9. There is no need to do calculations greater than the square root of P. The proof is really simple in retrospect and can be understood by an algebra student familiar with the law of quadratic reciprocity. I doubt that it could be simplified by using Class Field Theory. Perhaps it may be a trival application of class theory but I don't know. I skipped college after taking the equivalent of two years of courses via night school (none in higher algebra or number theory) and am now retired. However, my interest in number theory is such that I would be willing to pay to audit a few courses such as Algebraic Number Theory once I take care of a few other things like building my dream home. The book I purchased is mostly Greek to me.

My proof just got extended
Theorem
Let $$N = F_{x,y} = x^{2} + B*x*y -y^{2}$$ where gcd(x,y) = 1
Then there exists a coprime pair $$(x_{1},y_{1})$$ such that
$$F_{x_{1},y_{1} = N^{2^{m}}$$
If B = 0 then my proof relates to pythagorean triples
e.g.
$$2^{2} - 1^{2} = 3$$
$$5^{2} - 4^{2} = 3^{2}$$
$$41^{2}-40^{2}=3^{4}$$
$$3281^{2}-3280^{2}=3^{8}$$

$$3^{2}-2^{2}=5$$
$$13^{2}-12^{2}=5^{2}$$
$$313^{2}-312^{2}=5^{4}$$
$$195313^{2}-195312^{2}=5^{8}$$