Is There a Real Solution for a^n+b^n=c^n as n Approaches Infinity?

[Data]

There ceases to be a solution set {a,b,c} (alright, that is "independent" from n) of finite reals as n\rightarrow \infty.

What precisely do you mean by this statement? My reading is that you are trying to say that if \forall n \in \mathbb{N}^*, \ \mbox{we define} \ S_n = \{ (a, \ b, \ c) \in \mathbb{R}^3 | \ a^n+b^n=c^n \}, \ \mbox{then} \ \lim_{n \rightarrow \infty} S_n = \phi, which is silly, since \forall \ (a, \ b \in \mathbb{R}^+, \ n \in \mathbb{N}^*), \ \exists c\in \mathbb{R} \ \mbox{s.t.} \ a^n+b^n = c^n.

 

[Icebreaker]

Yes, but not if a,b,c are integers, as I explained above what I tried to do.

 

[Data]

Of course not. If we require a, \ b, \ c \in \mathbb{N}^* then there are no solutions \forall n \geq 3 by Fermat's Last Theorem. I was replying to

There is at least a solution set {a,b,c} of finite reals for every integer n>2. There ceases to be a solution set {a,b,c} (alright, that is "independent" from n) of finite reals as n \rightarrow \infty.

and

I wanted to know if I can extend this proof for all reals of a, b and c, which is why I posted it here anyway.

 

[matt grime]

I still think it is a bad question, since we can't seem to agree on the proper interpretation. Thank you for your time.

And here we do agree (or at least we can agree that the version of the question posted was ambiguously written). But I don't think this justified you jumping all over the OP with spurious comments about solving things "at infinity" or whatever the phrase was, espcially given your example using logs.

I am also more than confused as to what the OP wanted to even solve, especially if he was extending FLT in some way, since there are obvisouly solutions to a^n+b^n=c^n for all n integer with a,b,c real, and I thus may well have been miscrediting him with showing something he wasn't showing. (Yes, squaring both sides would be better, but proofs aren't unique. Oddly no one pointed out the flaw that you can't exchange limits and quotients in the manner he did)

 

[Curious3141]

And here we do agree (or at least we can agree that the version of the question posted was ambiguously written). But I don't think this justified you jumping all over the OP with spurious comments about solving things "at infinity" or whatever the phrase was, espcially given your example using logs.

I am also more than confused as to what the OP wanted to even solve, especially if he was extending FLT in some way, since there are obvisouly solutions to a^n+b^n=c^n for all n integer with a,b,c real, and I thus may well have been miscrediting him with showing something he wasn't showing. (Yes, squaring both sides would be better, but proofs aren't unique. Oddly no one pointed out the flaw that you can't exchange limits and quotients in the manner he did)

For the record, please note that I did not "jump all over the OP" until said OP started to evince disdain and rudeness by using phrases like "save your breath" and "get it through your head". I would like to think I am fairly civil until provoked.

When a person is wrong AND rude, I don't see the need for tact in return.

 

[matt grime]

Fair enough.

But you should realize it is perfectly possible to solve for things that have limits in them if you're going to criticize that.

For instance, I can "solve" lim (1+x/n)^n=2, the answer being, as you know, x=log2

 

[Curious3141]

Fair enough.

But you should realize it is perfectly possible to solve for things that have limits in them if you're going to criticize that.

For instance, I can "solve" lim (1+x/n)^n=2, the answer being, as you know, x=log2

Yes, that is fair enough. Does the RHS necessarily have to be a constant term (for this sort of "limit equation" to work), or can it be dependent on x as well ?

 

[matt grime]

I could put an x in the rhs if I felt like it since that "equaiton" above is exactly the same as

e^x=2,
so I could ask if there is a (real) x satisfying e^x=3x. (there is, but it won't be possible to find algebraically, I suspect)

 

[Curious3141]

I could put an x in the rhs if I felt like it since that "equaiton" above is exactly the same as

e^x=2,
so I could ask if there is a (real) x satisfying e^x=3x. (there is, but it won't be possible to find algebraically, I suspect)

The reason I'm asking is that (assuming the OP meant something like this), the original question is ambiguous because there're variable terms on each side.

The question can be rewritten in two ways :

1) "Solve" (n -> infinity)lim(a^n + b^n - c^n) = 0 for positive real (a,b,c).

In which case, there are no solutions for any a,b or c greater than or equal to unity. Every permutation of real (a,b,c) with each value less than one is a valid solution.

2) "Solve" (n -> infinity)lim(a^n + b^n)^(1/n) - c = 0 for positive real (a,b,c).

Here the solution is always c = max(a,b)

Both forms are algebraically equivalent, yet the solution set is different depending on which form is used. At least when the constant term is on the RHS, an unambiguous problem statement can be constructed.

 

[matt grime]

Both forms are not equivalent though because of that "limit" word in there (n is a dummy vairable, not actually a variable).

 

[Curious3141]

Both forms are not equivalent though because of that "limit" word in there (n is a dummy vairable, not actually a variable).

Yes, and this is the reason I had misgivings about solving equations with the limit notation in them. Certainly that works fine in special cases, as you illustrated.

 

[mathwonk]

this thread was so depressing from the beginning i never jumped in. questions need to be clearly stated before an answer is possible. the problem posed originally was never actually defined.

please stop this.

 

[Icebreaker]

If I could edit this, I would simply change "real" to "positive integer", and the problem would go away.