Proving Non-Equality of Cubes of Natural Numbers

[mathmaniac1]

Prove that
\(\displaystyle a^3+b^3 \ne to \ c^3 \ if \ a,b \ and \ c \in \ {N}\)

This is not a challenge,I am asking for help...

Any help is appreciated...

Thanks...

 

[MarkFL]

Re: a^3+b^3 not equal to c^3

By Fermat's Last Theorem, which was proved by Andrew Wiles, we know there are no solutions to the Diophantine equation:

\(\displaystyle x^n+y^n=z^n\)

for $2<n$.

 

[mathmaniac1]

Re: a^3+b^3 not equal to c^3

By Fermat's Last Theorem, which was proved by Andrew Wiles, we know there are no solutions to the Diophantine equation:

Yeah,thats what I am doing...trying to prove FLT...
And this is a first step,please tell me how can I prove this

 

[Nono713]

Re: a^3+b^3 not equal to c^3

There are standalone proofs for a few of the simpler exponents (3, 4, 5, ..) but the complete proof for all exponents is hardcore - you probably don't want to try and "prove FLT". Many mathematicians have tried and failed. Only one has succeeded and the mathematical background required is insane.

Take a look here to get some insight: Proof of Fermat's Last Theorem for specific exponents - Wikipedia, the free encyclopedia

 

[mathmaniac1]

Re: a^3+b^3 not equal to c^3

But I believe proving it for n=3 is simple...

Am I wrong...?

 

[Nono713]

Re: a^3+b^3 not equal to c^3

But I believe proving it for n=3 is simple...

Am I wrong...?

Not at all, proving it for n = 3 is not too difficult. But you are not "proving FLT" in this case, you are only proving one of infinitely many cases of the theorem.

 

[mathmaniac1]

Re: a^3+b^3 not equal to c^3

I know I cannot prove FLT for infinite,so I decided to try it for a few n.But I found it impossible(for me)to do...

Now can you give the proof,Bacterius?

 

[Nono713]

Re: a^3+b^3 not equal to c^3

I knew I cannot prove FLT for infinite,so I decided to try it for a few n.But I found it impossible(for me)to do...

Now can you give the proof,Bacterius?

I just linked it..

 

[mathmaniac1]

Re: a^3+b^3 not equal to c^3

Ok...

Proving it for any n requires knowledge on group theory and galois theory,right?

I will come back at it someday...

 

[MarkFL]

Re: a^3+b^3 not equal to c^3

If I recall correctly, Wiles essentially cut himself off from the world and worked on it in virtual solitude for seven years.

 

[mathmaniac1]

Like John Nash in "A Beautiful Mind"...??

 

[alyafey22]

Two days ago I watched a video about FLT here .It gave me some hope to try the Riemann Hypothesis . The interesting thing is that Wiles used an advanced method to prove it (using elliptic curves and modular forms) but Fermat claimed to have a proof for it , was it (if it exists) as complicated as Wiles's? , I don't think so !

 

[MarkFL]

I believe the consensus is that Fermat did not have a general proof as he claimed, but rather had a proof of a specific case or cases at best. I don't see how an elementary proof could have eluded the world's greatest mathematical minds for nearly 360 years.

 

[alyafey22]

I believe the consensus is that Fermat did not have a general proof as he claimed, but rather had a proof of a specific case or cases at best. I don't see how an elementary proof could have eluded the world's greatest mathematical minds for nearly 360 years.

Most of the time , it is not how elementary a proof is but rather how brilliantly we construct it .

 

[mathbalarka]

Proving it for any n requires knowledge on group theory and galois theory,right

Yes. It is basically based on the statement that every semistable elliptic curve is modular. It took my quite long to understand the whole proof of modularity theorem.

FLT for n = 3 is easy. I know two proofs of it, one by Fermat's Infinite Descent and another by usual arithmatic over \(\displaystyle \mathbb{Q} \left (\sqrt{3} \right ) \)

 

[caffeinemachine]

Re: a^3+b^3 not equal to c^3

I know I cannot prove FLT for infinite,so I decided to try it for a few n.But I found it impossible(for me)to do...

Now can you give the proof,Bacterius?

Hello mathmaniac,

Its a good thing that you are attempting to prove such things. But I'd like to mention something here. If you need help on such questions (and even on easier ones) I think you must show some effort. At least spend some time googling and many times that will solve your query. Post some of your ideas, good or bad, along with the question. Bacterius had linked the proof you were looking for but I guess you overlooked. Please understand that it takes a lot of effort and time to write a math proof in LaTeX.

I don't mean to upset anyone who helped out in this thread. My apologies in advance.

-Caffeinemachine.

 

[mathmaniac1]

Re: a^3+b^3 not equal to c^3

But I'd like to mention something here. If you need help on such questions (and even on easier ones) I think you must show some effort.

I thought a lot on it.And you mean show it here?Ok,I first thought about n=2 and why it works.
And I understood why it works.I don't think its any use to type it here...

Ok,let me ask another question.Forget the earlier one.

Any square number can be written as a sum of odd numbers.Can the cubes be written any any ways like it?

At least spend some time googling and many times that will solve your query.

Honestly,I didn't do it.I always see google as a last resort.(but I also use it for hints)
I was actually looking for a hint to try again to prove it on my own...
I thought typing \(\displaystyle a^3+b^3 \ne c^3\) in google won't give the results I wanted.But I never tried...

Post some of your ideas, good or bad, along with the question. Bacterius had linked the proof you were looking for but I guess you overlooked. Please understand that it takes a lot of effort and time to write a math proof in LaTeX.

Its okay,I can clearly understand the hardship of latexing,I've known it...

I don't mean to upset anyone who helped out in this thread.

I also think I should've used google rather than approaching here for a hint.

 

[mathmaniac1]

FLT for n=2(as I know it)

\(\displaystyle a^2+b^2\) can be equal to \(\displaystyle c^2\) in some cases as we know about Phythagorean triplets...

Any square number can be written as a sum of consecutive odd numbers.
4=1+3,9=1+3+5...etc

So \(\displaystyle a^2=1+3+5+7+9...\)
So \(\displaystyle a^2+b^2=(1+3+5+...) \ \ \ + \ (1+3+...)\)
Also \(\displaystyle c^2=1+3+5+...\)

This can happen if \(\displaystyle b^2\) adds to the consecutive string of odd numbers of \(\displaystyle a^2\) or vice versa...

So there can be \(\displaystyle a,b \ and \ c \ all \in N \) such that \(\displaystyle a^2+b^2=c^2\)

 

[mathbalarka]

FLT says there are no tuples (a,b,c) in R³ such that a^n + b^n = c^n for n > 2. There are infinitely many solutions for n = 2.

 

[mathmaniac1]

Yeah and I 've just shown why it is possible...

Can this be used to prove for n=3 and for any n(too ambitious)...?

 

[mathbalarka]

Absolutely not.

 

[mathmaniac1]

OK...

 

[alyafey22]

There are infinitely many solutions for n = 2.

How to prove they are infinite ?

 

[mathmaniac1]

Check my proof(not a perfect proof actually,better to be called reasoning)
If you can't see it tell me because I think I got it...

 

[alyafey22]

Check my proof(not a perfect proof actually,better to be called reasoning)
If you can't see it tell me because I think I got it...

I can't see how general is that

 

[alyafey22]

Let the following :

\(\displaystyle a^2=\sum^{n}_{k=0}2k+1 \,\,\, n>0\)

\(\displaystyle b^2=\sum^{m}_{k=n}2k+1 \,\,\, m-n>0\)

\(\displaystyle c= \sum^{n}_{k=0}2k+1\,+\, \sum^{m}_{k=n}2k+1=\sum^{m}_{k=0}2k+1\)

hence c must be a square number ...

 

[Nono713]

Let the following :

\(\displaystyle a^2=\sum^{n}_{k=0}2k+1 \,\,\, n>0\)

\(\displaystyle b^2=\sum^{m}_{k=n}2k+1 \,\,\, m-n>0\)

\(\displaystyle c= \sum^{n}_{k=0}2k+1\,+\, \sum^{m}_{k=n}2k+1=\sum^{m}_{k=0}2k+1\)

hence c must be a square number ...

Or just use $3^2 + 4^2 = 5^2$ and repeatedly multiply by four :) for any exponent $n$ one solution implies infinitely many just by multiplying each side by $m^n$ for $m > 1$.​

 

[MarkFL]

There's also Euclid's method for generating Pythagorean triples:

\(\displaystyle \left(m^2-n^2,2mn,m^2+n^2 \right)\)

where \(\displaystyle n<m\in\mathbb{N}\).

 

[mathbalarka]

How to prove they are infinite ?

It's quite trivial to prove that. As Bacterius suggested, we can use 3^2 + 4^2 = 5^2 and multiply it by square numbers to reckon that they are actually infinite. In fact, the number of Primitive Pythagorean triples smaller than a given number n is proportional to n.

 

[mathbalarka]

The number of Primitive Pythagorean triples smaller than a given number n is proportional to n.

This result can be generalized further :
If we denote \(\displaystyle P(N)\) as the number of PPTs less than N, we have

\(\displaystyle P(N) \sim \frac{N}{2\pi}\)

Balarka
.

 

[mathmaniac1]

It's quite trivial to prove that.

Absolutely!My understanding

Let \(\displaystyle b^2\) be an odd square number.There exists a pythagorean triple including \(\displaystyle b^2\) because the odd number \(\displaystyle b^2\) can form the last number of the string of consecutive odd numbers of another square number \(\displaystyle a^2\).So there are PPTs as long as there are odd numbers and their squares,so they are infinite...

 

[mathworker]

i have a little different approach to prove there are infinite triples ,(though its unnecessary)
let x,z be two variables ,
let x^2+z^2=(x+y)^2
then z^2=y^2+2xy =y(y+2x)
and if we consider y to be perfect square we can adjust x such that y+2x is perfect square
since there are infinite perfect squares there and infinite x's and thus we can find infinite z's thus there are infinite solutions to x^2+y^2=z^2

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i tried applying same method to n=3 but i found myself cycling around ...