# A problem like Fermat's Last Theorem

1. Mar 11, 2006

### kmarinas86

A "problem" like Fermat's Last "Theorem"

Given the equation:

$a^n+b^n=c^n$
Where:
a, b, and c are integers.
And n is real number greater than 2.

Is there a solution?

edit: a, b, and c are supposed to be non-zero like in Fermat's Last Theorem

Last edited: Mar 11, 2006
2. Mar 11, 2006

### Hurkyl

Staff Emeritus
Yes. Try picking a, b, and c before you worry about n.

3. Mar 11, 2006

### kmarinas86

But a, b, and c are supposed to be non-zero integers. That's what Fermat's Last Theorem is about.

4. Mar 12, 2006

### Hurkyl

Staff Emeritus
Okay, pick them to be nonzero integers. Then worry about n.

5. Mar 12, 2006

### HallsofIvy

Staff Emeritus
You asked whether there were any positive integers a, b, c, such that, for some n, an+ bjn= cn. Hurkyl's suggestion was that you choose suitable a, b, c, then look for a value of n.

6. Mar 12, 2006

### kmarinas86

That doesn't answer my question though.

7. Mar 12, 2006

### arildno

As your first task, let a and b be the same integer.

8. Mar 12, 2006

### Muzza

Woe is you. These people sure are mean when they don't tell you the solution straight away.

9. Mar 12, 2006

### kmarinas86

Fermat's Last Theorem goes like this:

Where a, b, c, and n are all integers:
And where n is an integer greater than 2,

There can be no a, b, c, n, where:
$a^n+b^n=c^n$

What non integer n should I use then? There are literally an infinite amount. I would choose n first because it was already proven that the formula does not work where n is an integer.

Do you understand that it took hundreds of years to prove Fermat's Last Theorem? It looks so simple, but it's not. By experience it is known by mathematicians that there is no such thing as a simple proof "thus far" of Fermat's Last Theorem. Therefore, the answers I have gotten so far have been inadequate.

I haven't found any substantation whatsoever whether or not the no solution answer (which was proven for the n=integer case) applies for the n=real number case.

Last edited: Mar 12, 2006
10. Mar 12, 2006

### matt grime

Of course it doesn't apply. And I think those people who posted here to help you are more than familiar with the history of FLT, and quite a few might even know about Wiles's proof.

Fix a<b<c and you're asking if a^n+b^n-c^n= 0. Try and find some simple conditions to ensure that there is some n where this happens. Hint: intermediate value theorem, and try to be polite to other people who are trying to help you to help yourself, it really is quite straight forward to find infinitely many a,b,c and n real and greater than 2, as a really big hint try a=b=1 and *any* c larger than 2 for an idea (if you want n>2 then try a little harder, though why n=1.5 won't do I'mi not sure).

NB fixing n can't help you at all, and you are after all trying to show that there is some n for which there are integer solutions, and as you observe there are infinitely many to choose from, so don't choose it first.

Last edited: Mar 12, 2006
11. Mar 12, 2006

### kmarinas86

1^2.5 < 1^2.5+1^2.5 < 2^2.5
1^3.5 < 1^3.5+1^3.5 < 2^3.5

In general
1^x < 1^x+1^x < 2^x
for any x greater than 1
where
1 < 2 < 2^x

There is no integer between 1 and 2.

Where x>2:
1^x+1^x < 2^x
1^x+1^x > 1^x

For any:
a^n+b^n=c^n
Where a, b, and c are integers, and n is a real number.
the nth root of (a^n+b^n) must equal an integer.

Last edited: Mar 12, 2006
12. Mar 12, 2006

### matt grime

Sorry, what does that show?

It is easy to find infinitely many (strictly positive) integer a,b,c and real n>2 with a^n+b^n=c^n using the intermediate value theorem with, so try doing it

13. Mar 12, 2006

### kmarinas86

It means:

This proves that for the equation:
1^x+1^x = c^x, where x>1
c cannot be an integer. It must be a number between 1 and 2. Of course, this is not a formal proof, but a formal proof can be derived from this. In fact, c must be the xth root of 2.
because
1^x+1^x = 2

Also
Any root of 2 cannot ever be less than 1.
And 1^x = 1.
And 2 to any xth root cannot ever be an integer. (for any x > 1)

You're right:

If you look at the graph below, you'll see that for x>2 as x approaches infinity, the graph line becomes asymptotic. Soon I will post a graph that has a bunch of lines, maybe dozens, that will probably show the same thing. But I have found, that indeed, there are solutions.

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Last edited: Mar 12, 2006
14. Mar 12, 2006

### matt grime

It is very easy to do (to find infinitely many a,b,c integers and n a real larger than 2). We are not lying to you, nor are we pretending not to know the answer when we don't. Do you know what the intermediate value theorem states? That is all that is required, and a little thought (the case a=b=1 surely tells you something, how about making a=b=4?), and infinitely many solutions can be shown to exist by anyone who's done a first course in analysis.

Instead of assuming that we're all wrong, why don't you possibly consider that we might be right and try following the hints you've been given? You'll learn more by doing so. As I say, infinitely many examples can be shown to exist by anyone with a first course in analysis behind them.

Let me help even more by pointing out that n=log(1/2)/log(4/5) which is greater than two will have integer solutions to a^n+b^n=c^n.

Indeed, now I come to think of it anyone with highschool maths can create arbitrarily many examples. A first course in analysis will explain 'why' this is possible. As I say, it is merely the intermediate value theorem.

15. Mar 12, 2006

### kmarinas86

Solution for 3^x+3^x=4^x:

ln(2)/ln(4/3)=x

How to get it:
3^x+3^x=4^x
2*3^x=4^x
2=(4/3)^x
ln(2)=ln(4/3)*x
ln(2)/ln(4/3)=x

a, b, c are all integers, and x is a real number.

Mystery solved

#### Attached Files:

• ###### continuousfunction2.jpg
File size:
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16. Mar 12, 2006

### kmarinas86

It's ok, I figured out on my own before I saw this post