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

[Icebreaker]

I'm trying to show the following:

Let a^n+b^n=c^n where a, b, c are not equal to zero. There is no real solution for a, b, c as n\rightarrow\infty.

Proof:

Assume the contrary. If there is a real solution for a, b, c as n\rightarrow\infty, then

lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=1

must also be true.

We can use properties of limits and algebra to obtain, from the previous equation

lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1

It can be shown that if a^n+b^n=c^n (where a, b, c are not equal to zero) is true, then c>a and c>b

Therefore

0<\frac{a}{c}<1 and 0<\frac{b}{c}<1

Using properties of limits, we can state that

lim_{n\rightarrow\infty} \frac{a^n}{c^n} = lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 0

and therefore

lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=0

Which is not 1, and we have thus shown that the contrary to the first statement cannot be true.

p.s. I hate tex

 

[robert Ihnot]

Icebreaker,
lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1

for starters where did you get this?

Before we get too caught up in that, I have to ask, What does a limit have to do with it?

 

[Icebreaker]

lim_{n\rightarrow\infty}\frac{a^n+b^n}{c^n}<br /> =lim_{n\rightarrow\infty}\frac{a^n}{c^n}+\frac{b^n}{c^n}<br /> =lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}<br /> =1

If you have a way to prove it without limits, be my guest.

 

[Curious3141]

First of all, I don't like the problem "statement" at all, it makes little mathematical sense. Did you get that out of a textbook, or did you make it up ?

The ("Fermat-like") equation a^n + b^n = c^n has infinitely many real solution sets (a,b,c) for every positive integral value of n. Asking what solutions exist when n is infinite, or even "n tends to infinity" is meaningless. This is not evaluating the limit of an expression, this is gobbledegook.

 

[Icebreaker]

Why is asking what solutions exist when n increases without bound meaningless?

 

[Curious3141]

Think about what you're asking.

Do real solution sets (a,b,c) exist for n = 1000 ? YES

Do real solution sets (a,b,c) exist for n = 100,000 ? YES

Do real solution sets (a,b,c) exist for n = 10^(10^(10^10) ? YES

Is n "tending" to infinity in the above ? I suppose so.

Do real solutions exist all the way regardless of the finite value of n, no matter how large ? Obviously.

Does it make sense to ask the question for an infinite value of n ? NO.

***

Do you see now ?

 

[Icebreaker]

Explain what mathematical principal I'm violating here? Because from what I see above I have every right asking the question I did.

 

[Curious3141]

Explain what mathematical principal I'm violating here? Because from what I see above I have every right asking the question I did.

As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large. Note that even if the limit exists, the expression can NEVER actually equal the limit. It can only approach it.

Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.

 

[Icebreaker]

As has been discussed a very large number (but still finite) number of times in this forum, infinity is NOT a number. You cannot set up an equation using infinity and expect it to have a solution.

Oh? lim_{n\rightarrow\infty}\frac{1}{n}=0

If, OTOH, you're asking whether that equation has solutions for very large finite values of n, then the answer is always YES, so the premise you want to prove is wrong.

Of course there's a solution for every finite n. So?

When you are evaluating the limits of expressions as something "tends" to infinity, all you are doing is investigating the behaviour or a mathematical expression as one of the terms is made arbitrarily large.

So I'm investigating the value of the expression as a variable is increasing without bounds. And?

Just as you cannot speak of the expression ever having the ACTUAL value of the limit, you cannot set up an equation using those rules.

I never said n EQUALS infinity.

 

[Icebreaker]

Think about what you're asking.

Do real solution sets (a,b,c) exist for n = 1000 ? YES

Do real solution sets (a,b,c) exist for n = 100,000 ? YES

Do real solution sets (a,b,c) exist for n = 10^(10^(10^10) ? YES

Congratulations, you've proven that there are at least 3 solutions for the equation.

Is n "tending" to infinity in the above ? I suppose so.

Why do you think I set up the limit? Because I don't simply "suppose so".

Do real solutions exist all the way regardless of the finite value of n, no matter how large? Obviously.

I never said they didn't.

Does it make sense to ask the question for an infinite value of n ? NO.

And here is where your logic breaks down.

 

[Icebreaker]

By your logic, suppose we have y=\frac{1}{x}. We find that there is a non-zero y for any arbitrarily large x, and therefore makes no sense to ask whether there's a limit for y=\frac{1}{x} as x\rightarrow\infty.

 

[Justin Lazear]

I never said n EQUALS infinity.

Does it make sense to ask the question for an infinite value of n ? NO.

And here is where your logic breaks down.

I'm confused... Do you want n to equal infinity or not?

--J

 

[Icebreaker]

No, I don't. Read the post above yours.

 

[DeadWolfe]

Could you state your meaning using a more rigorous definition of limits?

 

[robert Ihnot]

To Illustrate: a case where you could use the limit is the case of the Fibonacci numbers: 1,1,2,3,5,8,13,21,34,55,89,144...where the series in continued by the formula: F(N-1)+F(N)=F(N+1), or we add two successive numbers to get the next one.

Now it happens to be that this series is closely connected with the Golden Mean:
\frac{-1+\sqrt(5)}{2} =.618034...

We look at: 34/55 =.618182..;55/89 =.617978...;89/144 =.618056...

Thus we might wonder if the limit as n goes to infinity of F(n)/F(n+1) = Golden Mean...which it does!

This limit is so good, that if divided 144 by the Golden Mean we get 232.996, which rounds off to 233 =89+144. (In fact it is good even from the second "1" giving 1/GM = 1.6180 rounds to 2. 2/GM =3.23 rounds to 3, 3/GM=4.85 rounds to 5!

 

[Justin Lazear]

Do real solutions exist all the way regardless of the finite value of n, no matter how large? Obviously.

I never said they didn't.

I'm trying to show the following:

Let a^n+b^n=c^n where a, b, c are not equal to zero. There is no real solution for a, b, c as n\rightarrow\infty.

And I don't understand what the post that supposedly explains whether or not you want n to be infinity. I'm still confused on this issue.

--J

 

[Curious3141]

Oh? lim_{n\rightarrow\infty}\frac{1}{n}=0

Oh-what ? \frac{1}{n} is an EXPRESSION, not an EQUATION. And it *never* equals zero no matter what value of n you choose. Only the limit is zero. There is a fundamental distinction here you don't seem to be grasping.

Of course there's a solution for every finite n. So?

So the premise you set about to "prove" is WRONG !

I'm investigating the value of the expression as a variable is increasing without bounds. And?

No, you are NOT. You are investigating an EQUATION as n tends to infinity. Which makes no sense.

You CAN investigate the behavior of the EXPRESSION {(a^n + b^n)}^{\frac{1}{n}} as n tends to infinity. If, for simplicity, we allow only for positive real values of a and b, then

\lim_{n \rightarrow \infty}{(a^n + b^n)}^{\frac{1}{n}} = max(a,b)

Fair enough ?

*If* you say c = {(a^n + b^n)}^{\frac{1}{n}} THEN it's perfectly acceptable to say :

\lim_{n \rightarrow \infty}c = max(a,b)

Here you are still investigating the behavior of the expression called "c" as n tends to infinity. It's completely fine.

OTOH, it is not standard practice (and hence not acceptable) to do the same for an equation trying to find solutions at a limit.

 

[DeadWolfe]

I tihnk your problem is that you haven't defined "solution" well enough, when you said there wasn't one.

 

[Icebreaker]

Oh-what ? \frac{1}{n} is an EXPRESSION, not an EQUATION.

Oh? And

\frac{a^n+b^n}{c^n}

is NOT an expression?

So the premise you set about to "prove" is WRONG !

Interesting how you can rush to that conclusion based on 3 examples where n equalled finite numbers.

No, you are NOT. You are investigating an EQUATION as n tends to infinity. Which makes no sense.[...]

No, I am investigating the EXPRESSION

\frac{a^n+b^n}{c^n}

as n increases without bound. And from what I see, there's little difference between what I am trying to do, and the example which you gave.

 

[Icebreaker]

I tihnk your problem is that you haven't defined "solution" well enough, when you said there wasn't one.

Perhaps. I will try and find a better definition of "solution".

And I don't understand what the post that supposedly explains whether or not you want n to be infinity. I'm still confused on this issue.

--J

No, you read the wrong post. Read the one directly above your first reply to this thread.

 

[Curious3141]

Let's take things one at a time...

Oh? And

\frac{a^n+b^n}{c^n}

is NOT an expression?

This IS an expression. If you have rigidly defined c^n = a^n + b^n then the expression is ALWAYS equal to unity. The limit is also unity.

Going back to your first post :

lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=1

must also be true.[/tex]

Fine so far.

We can use properties of limits and algebra to obtain, from the previous equation

lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=1

Also fine.

It can be shown that if a^n+b^n=c^n (where a, b, c are not equal to zero) is true, then c&gt;a and c&gt;b

True for all finite n, but NOT at the limit. At the limit, c = max(a,b). If b is the greater, then c = b at the limit. This is your slip-up.

Therefore

0&lt;\frac{a}{c}&lt;1 and 0&lt;\frac{b}{c}&lt;1

Again, true for all finite n cases, not correct at the limit.

Using properties of limits, we can state that

lim_{n\rightarrow\infty} \frac{a^n}{c^n} = lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 0

Incorrect ! If b is the larger quantity, then lim_{n\rightarrow\infty} \frac{a^n}{c^n} = 0 while lim_{n\rightarrow\infty} \frac{b^n}{c^n} = 1

The rest is all wrong too.

and therefore

lim_{n\rightarrow\infty} \frac{a^n}{c^n}+lim_{n\rightarrow\infty} \frac{a^n}{c^n}=lim_{n\rightarrow\infty} \frac{a^n+b^n}{c^n}=0

Which is not 1, and we have thus shown that the contrary to the first statement cannot be true.

Interesting how you can rush to that conclusion based on 3 examples where n equalled finite numbers.

I'm sorry, but you do not understand limits.

 

[Curious3141]

You could have saved a lot of breath: my mistake was not in the way I am evaluating the limits, nor my understanding of limits, but circular logic. I assumed n\rightarrow\infty, but used a finite number to evaluate one of the properties I used in the proof, and replugged it back into my original guess.

Saved a lot of breath ? How about next time I don't bother to correct your misconceptions at all ?

And there IS something deeply wrong with your understanding of limits if you can make a mistake like that.

I'm done with this thread.

 

[Icebreaker]

Get it through your head: limits had nothing to do with it. It was circular logic.

 

[Curious3141]

Get it through your head: limits had nothing to do with it. It was circular logic.

GET IT THROUGH YOUR HEAD : Your entire attempt to find "solutions" to an "equation" at the limit was FLAWED. It wasn't a simple case of mistaking the behaviour of a finite property with the limit, IT WAS A COMPLETELY FLAWED ATTEMPT AT A PROOF BY CONTRADICTION !

FUNDAMENTALLY WRONG CONCEPT ! Not "circular logic".

 

[Icebreaker]

My entire attempt to find "solutions" to the limits of both sides of the "equation" by contradiction was FLAWED because I mistook the behavior of a finite property with the limit.

Tell me what's wrong with the following, by your definition

x = x

lim_{x\rightarrow\infty}x = lim_{x\rightarrow\infty}x

 

[chronon]

Icebreaker's proof is perfectly valid. But all it shows is that there are only a finite number of values of n for which a^n+b^n=c^n for fixed a,b and c >0.

 

[matt grime]

I want to echo Chronon because Icebreaker was getting a lot of stick for someone else misunderstanding the question.

Rephrase it as consider the function f_n = x^n +y^n

and let f be the pointwise limit of f_n when this makes sense.
Are there any pairs x,y for which f(x,y)=1?

No, because the pointwise limit is identically zero for 0<x,y<1, 2 for x=y=1, and not defined other wise.

 

[Curious3141]

I want to echo Chronon because Icebreaker was getting a lot of stick for someone else misunderstanding the question.

Rephrase it as consider the function f_n = x^n +y^n

and let f be the pointwise limit of f_n when this makes sense.
Are there any pairs x,y for which f(x,y)=1?

No, because the pointwise limit is identically zero for 0<x,y<1, 2 for x=y=1, and not defined other wise.

Matt, look at the original question and the way it was phrased. Does it make sense ?

It was not the limit of an expression. It was an attempt to define real solutions to an *equation* with exponents as the exponents tended to infinity. As far as I'm aware, that is not sound mathematics.

Are you saying I'm wrong ? If so, please explain why.

 

[matt grime]

Yes, the original question makes sense as in it can certainly be understood even if it isn't the clearest way of writing it out: it defines a function as a pointwise limit. It would ghave been better to explain how it should have been better written since it is obvious what it is asking for.

How can you simultaneously say it isn't a limit and that things tend to infinity?

If you prefer, it is asking for the direct limit of the sets of solutions for each n ordered by inclusion - this makes it even clearer the answer is empty.

 

[Curious3141]

Yes, the original question makes sense as in it can certainly be understood even if it isn't the clearest way of writing it out: it defines a function as a pointwise limit. It would ghave been better to explain how it should have been better written since it is obvious what it is asking for.

No, it wasn't obvious to me what the question was asking for.

For me the solution to an equation has to satisfy the equation when it is substituted back into it. Agree ?

Now let me explain my problem with the question. If one tries to solve the equation algebraically "at the limit", the solution might just not work on the original equation.

Let's say we decide the values of a and b and want to determine what value of c (if any) exists. The obvious way of doing that is to evaluate c = {(a^n + b^n)}^{\frac{1}{n}}. It is clear that as n tends to infinity, the limit of c = max(a,b). Fair enough ?

Now let's try putting that back into the original equation in the form a^n + b^n - c^n = 0, where n tends to infinity. Here we are evaluating each term sequentially. It is clear that no solution set other than the trivial (0,0,0) will actually work upon substitution.

I believe the original poster was trying to use this sort of logic to show that the equation can have no real solutions "at the limit". But to me, the question makes no sense. If you solve an equation with valid algebra, the solution *has* to work. Putting the solutions back in and verifying that they do not satisfy the equation does not mean that there are no real solutions to the equation per se, it means that the statement itself is flawed.

The equation (barring the limit) is in a valid algebraic form. The solutions that I found were with valid algebra. The problem was in the way that solution behaved when the limit was taken. That limit could not successfully be substituted back into the original equation.

The conclusion I draw from that is that it *makes no sense* to speak of solving an equation at the limit. It's fine to define the behaviour of a function at the limit, as you have done, but not try to solve an equation.

Do you still disagree, Matt ?

 

[Icebreaker]

Why does it "make no sense" to solve an equation at the limit?

x = x

lim_{x\rightarrow a}x = lim_{x\rightarrow a}x = a

Maybe you can explain your definition of "solving an equation" better.

 

[Curious3141]

Why does it "make no sense" to solve an equation at the limit?

x = x

lim_{x\rightarrow a}x = lim_{x\rightarrow a}x = a

Maybe you can explain your definition of "solving an equation" better.

Your notation is valid for finite a, but it breaks down for infinite a. In the case of a simple identity, the break-down is not obvious.

Try this :

lim_{x \rightarrow \infty}(x - 1) = lim_{x \rightarrow \infty}x

Does that imply that you've solved x -1 = x ?

EDIT : Made it a much simpler example.

 

[Icebreaker]

No, it implies that as x\rightarrow\infty, the two expressions are equal. Without the infinite limit condition, the statement "the two expressions are equal" is false. Otherwise, it is true.

In other words, x - 1 = x is true iff x\rightarrow\infty.

Or, if you prefer:

f(x)=f(x) \Rightarrow lim_{x\rightarrow\infty}f(x) = lim_{y\rightarrow\infty}f(y)

However

lim_{x\rightarrow\infty}f(x) = lim_{y\rightarrow\infty}f(y) does not imply that f(x)=f(x).

 

[Curious3141]

No, it implies that as x\rightarrow\infty, the two expressions are equal. Without the infinite limit condition, the statement "the two expressions are equal" is false. Otherwise, it is true.

I would prefer to say the expressions approach one another, and the limits of the expressions are equal. Now think : is that really an equation ? Have you truly "solved" for a real value of x ? Real numbers have to be finite.

By the same token, in your original problem statement, any real number set (a,b,c) that's put into the equation "at the limit" will be either reduced to zero or be increased without limit. From that you think you can say that "there is no real solution" to the equation at the limit (which is what I think you tried to prove).

But in truth, there are no real solutions to *any* (non-trivial) equations at infinite limits. This is why I feel that it makes no sense to even ask the question.

I just saw your edit :

In other words, x - 1 = x is true iff x\rightarrow\infty

I disagree with this statement. I think you should phrase that as "the limit of x-1 is equal to the limit of x as x tends to infinity). Subtle difference, semantics perhaps, but to me, it makes a lot of difference.

 

[Icebreaker]

But in truth, there are no real solutions to *any* (non-trivial) equations at infinite limits. This is why I feel that it makes no sense to even ask the question.

What kind non-triviality are we talking about here? Because, as I pointed out before:

lim_{x\rightarrow\infty}\frac{1}{x}=a=0

And I disagree, it makes perfect sense to ask the question, even if you are right. Redundant, perhaps, but nothing illegal or illogical. After all, if you are right for any general case, then I can prove it for any special case, and I did. So what's the problem here?

 

[matt grime]

No, it wasn't obvious to me what the question was asking for.

well it was to me.

For me the solution to an equation has to satisfy the equation when it is substituted back into it. Agree ?

but there is no explicit equation to solve, so this is not relevant. there is a pointwise defined function.

Now let me explain my problem with the question. If one tries to solve the equation algebraically "at the limit", the solution might just not work on the original equation.

one doesn't solve it at the limit, one shows that no real a,b,c all strictly positive exist such that the limits agree and are not infinite, whether or not there is a solution for any particular n is not important, what is important is finding if there are real strictly positive a,b,c such that lim a^n + b^n =lim c^n with the limits taken as n tends to infinity (when it exists)

Let's say we decide the values of a and b and want to determine what value of c (if any) exists. The obvious way of doing that is to evaluate c = {(a^n + b^n)}^{\frac{1}{n}}. It is clear that as n tends to infinity, the limit of c = max(a,b). Fair enough ?

yes, and if you evaluate that limit what do you get?

The rest of you post isn't important, evaluating "sequentially? what does that even mean?


[/quote]Now let's try putting that back into the original equation in the form a^n + b^n - c^n = 0, where n tends to infinity. Here we are evaluating each term sequentially. It is clear that no solution set other than the trivial (0,0,0) will actually work upon substitution.

I believe the original poster was trying to use this sort of logic to show that the equation can have no real solutions "at the limit". But to me, the question makes no sense. If you solve an equation with valid algebra, the solution *has* to work. Putting the solutions back in and verifying that they do not satisfy the equation does not mean that there are no real solutions to the equation per se, it means that the statement itself is flawed.

The equation (barring the limit) is in a valid algebraic form. The solutions that I found were with valid algebra. The problem was in the way that solution behaved when the limit was taken. That limit could not successfully be substituted back into the original equation.

The conclusion I draw from that is that it *makes no sense* to speak of solving an equation at the limit. It's fine to define the behaviour of a function at the limit, as you have done, but not try to solve an equation.

Do you still disagree, Matt ?[/QUOTE]


we aren't solving at the limit, i don't see the original post using the phrase "at the limit"; only you appear to be using it. solve isn't the word I'd've used, but that is neither here nor there really. translating the question to one where we do not "solve" isn't very demanding

 

[Curious3141]

Sorry to dredge this up again, but I would like this case resolved :

Is there a stipulation against a, b, or c containing a term in n ?

What about a = {(1 + \frac{1}{n})}^{\ln{(e - 1)}}, b = 1 and c = (1 + \frac{1}{n}) ?

Those are all definitely real numbers. Put them into the equation and the limit simply becomes :

(e - 1) + 1 = e

So why isn't that a real solution set ?

EDIT : Although I don't know if a and c really qualify as "standard" real numbers, but that's why I'm asking the question.

 

[matt grime]

Because they aren't solutions for all n (or indeed any n).

Any three numbers of abs value less than 1 satisfy

lim(a^n+b^n)=lim(c^n)

but the question was asking what are the solutions for all n. The first line of the question stated that

a^n+b^=c^n.

And a and c aren't fixed real numbers, either, since they vary with n.

 

[Curious3141]

Because they aren't solutions for all n (or indeed any n).

Any three numbers of abs value less than 1 satisfy

lim(a^n+b^n)=lim(c^n)

but the question was asking what are the solutions for all n. The first line of the question stated that

a^n+b^=c^n.

And a and c aren't fixed real numbers, either, since they vary with n.

How do you mean "for all n" ? The original question didn't say "for all n" ?

Do you mean : does a particular positive real solution set (a,b,c) exist such that a^n + b^n = c^n holds identically for all natural n ? Because that can immediately be shown to be untrue - just square the thing.

I still am not sure what the question is asking. Certainly there're infinitely many real solutions (a,b,c) for each and every finite value of n. I'm still thrown by the use of the limit notation in the question.

But given the limit notation, I don't see why my solutions (with a and c in terms of n) cannot hold. Please explain this better.

 

[matt grime]

i guess this is the difference in the way we read the question.

 

[Curious3141]

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

 

[Icebreaker]

I still think it is a bad question

Well, I respectfully disagree.

 

[Curious3141]

Well, I respectfully disagree.

Then please explain exactly what you meant by the question. Does my solution set in terms of n work ? Your question said nothing about independence from n for the solutions.

 

[Icebreaker]

I deemed it unnecessary to state this "independence" because a violation of such "independence" would require the solution set to hold true for all n, otherwise it would not satisfy the equation a^n+b^n=c^n in the first place. Thus, the solution I want are not variables that change with n, but fixed numbers.

 

[Curious3141]

I deemed it unnecessary to state this "independence" because a violation of such "independence" would require the solution set to hold true for all n, otherwise it would not satisfy the equation a^n+b^n=c^n in the first place. Thus, the solution I want are not variables that change with n, but fixed numbers.

I'm very confused now. What sort of solution set are you looking for ?

1) Did you mean to imply that a particular fixed solution set (a,b,c) must satisfy the equation a^n + b^n = c^n (without the limit notation) as the value of n is allowed to vary ? For example, if (a,b,c) are positive reals that satisfy the equation when n = 2 then the same values will satisfy the same equation when n = 4 ? As I said, this is trivially falsifiable, just by squaring the LHS and RHS of the equation.

2) Did you mean that at least one solution set (a,b,c) can be found for the equation a^n + b^n = c^n for each and every natural number value of n ? Then I heartily agree, in fact, there are an infinite number of solutions for each and every such value.

3) Did you mean are there values (a,b,c) that fail to satisfy a^n + b^n = c^n for all natural numbers n, yet will satisfy the same equation when n is treated as infinite ? Because, as I have said c = max(a,b) is a solution set when either one or both of (a,b) is one or greater. If both (a,b) are less than one, then *all* permutations of (a,b,c) with 0 <a,b,c <1 will satisfy. With no further clarification of the question, I don't see why the other solution in terms of n and e cannot be a solution set as well.

4) Did you mean something else that I am completely missing ?

 

[Icebreaker]

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.

 

[Curious3141]

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.

In that case, the problem is equivalent to the case 3 that I mentioned, and solutions exist (meaning what you set about to "prove" is wrong). Here is the solution set :

(a, b, max[a,b]) is a solution set when either one or both of a, b is one or greater. *All* permutations of (a,b,c) with 0 < a,b,c <1 will satisfy.

 

[Icebreaker]

Then I will impose a restriction on the solution, where {a, b, c} must be integers which are greater than zero; thus reverting back to a special case of FLT, which was what I was set out to prove anyway.

 

[Curious3141]

Then I will impose a restriction on the solution, where {a, b, c} must be integers which are greater than zero; thus reverting back to a special case of FLT, which was what I was set out to prove anyway.

Sigh... you keep changing the question. Are you acknowledging that you were on the wrong track to begin with ?

If it's FLT, you don't need the limit or anything. FLT has already been proven.

 

[Icebreaker]

Actually this is the first time that I changed the question.

I know that a general proof of FLT had already been found; however, that doesn't prohibit me from proving special cases of it. In my original proof, I limited a, b, c and n to integers. 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.