Extended Fermat's last theorem.

[suyver]

"Extended" Fermat's last theorem.

Just to satisfy my own curiousity:

FLT states that there are no $n\in{\mathbb N}$ such that

$$x^n+y^n=z^n$$

whenever $n\geq 3$ and $x,y,z\in{\mathbb N}$.

However, what would happen when I allow $n$ to be non-integer as well? Are there solutions if $n\in{\mathbb Q}^+$ or $n\in{\mathbb R}^+$ ? Will one be able to find a set $x,y,z\in{\mathbb N}$ and an $n\geq 3$ such that this "extended" FLT holds?

 

[matt grime]

i think it is true that in general:
If x,y, and z are in N and p/q is a rational q not 1 and in lowest common terms then if x,y,z are coprime and not q'th powers the p/qth powers are linearly independent over Z.

proof in a simple example
if sqrt(x)+sqrt(y)=sqrt(z), then squaring both sides, sqrt(xy) is rational, which can't happen unless xy is a perfect square, contradicting the assumptions on 2nd powers and coprimality. of course if the numbers are 2nd powers then the question reduces to the usual fermat last theorem statement. the proof in the more general setting is messier.

there are of course trival cases sastifying your hypothesis such as 4,4,16 and 1/2 for x,y,z,n resp.

 

[suyver]

Originally posted by matt grime
i think it is true that in general:
If x,y, and z are in N and p/q is a rational q not 1 and in lowest common terms then if x,y,z are coprime and not q'th powers the p/qth powers are linearly independent over Z.

proof in a simple example
if sqrt(x)+sqrt(y)=sqrt(z), then squaring both sides, sqrt(xy) is rational, which can't happen unless xy is a perfect square, contradicting the assumptions on 2nd powers and coprimality. of course if the numbers are 2nd powers then the question reduces to the usual fermat last theorem statement. the proof in the more general setting is messier.

I'm still thinking about this part. It's quite hard for a non-mathematician... It seems to work for powers less than 1, but that wasn't my question (see below).

Originally posted by matt grime
there are of course trival cases sastifying your hypothesis such as 4,4,16 and 1/2 for x,y,z,n resp.

But I specifically stated $n\geq 3$, so your case does not apply as you have $n=1/2$...

Edit: the only example that I could think of quickly, was x=y=z=1 and any n, but that's "cheating". So let's restrict x,y,z to be strictly larger than 1.

 

[matt grime]

apologies for the incorrect counter example. but you did state fo n in Q+ or R too.


crappy argument removed


it suffices to consider only the case 1/q since if x,y,z are solutions for p/q then replacing them with x^p etc gives a solution for 1/q and the coprimality etc is preserved as p/q is in lowest terms. and there the only solutions are for qth powers, which implies a non-trival solution for u^p+v^p=w^p, which in turn implies p is 1 or 2.

 

[suyver]

OK, I think I understand your argument. Very beautifully done!

Thus, the result is that, for all $n\in{\mathbb Q},\; n\geq 3$:

$$x^n+y^n\neq z^n$$

if $x,y,z\in{\mathbb N},\; x,y,z>1$.



One final question: you've convinced me of the case where I can write $n=p/q$, but what of the case where $n\in{\mathbb R},\; n\geq 3$? Does your argument hold there too?

 

[matt grime]

I've got nothing on the general case for n a real number what so ever.

I think it is possible to find solutions - here's me reason

let x=y=2 and let n,z be arbitrary

2*2^n=x^n

take log_2

n+1=nlog_2(x)

if you let x=3, then I the resulting n satisfying that equation might be large enough for your needs, but I've not worked it out.

If it doesn't it at least tells you there are tricks you can pull to try and get an answer,and once you're allowed arbitrary n taking logs is allowed.

I've not proved the qth roots of coprime numbers that aren't qth powers are linearly independent over Z, but I'm very sure they are.

 

[matt grime]

I had this thought over dinner.

Let me switch to different letters to make it more suggestive, and hopefully easier to understand.

fix p,q,r positive integers

consider the function

x ---> p^x+q^x-r^xfrom R to R.

evaluate at x=3.

if there is another x greater than three which has the opposite sign to the value at hree then it follows there is some x greater than 3 satisfying your requirements.

I think that by choosing p,q,r well you will find a positive answer.say p=q=4 r=5, then at three you get

64+64-125 >0

at 4 you get 256+256-625<0

so there is a point where the graph cuts the axis between x=3 and 4edit, actually it's easier to consider

(p/r)^x+(q/r)^x

where p,q and r are chosen so that the quantity above is greater than 1 at x=3, and then as x goes to infinity,as long as p,q<r the quantity goes to zero, hence there is some x where it equals 1.

 

[suyver]

Yes, that makes sense. I think it's really cool how you just prove that a solution must exist, without bothering to calculate a particular solution. That's clearly the intelligent way!

Thank you for taking the time to explain this to me in such detail! (I didn't need the result for my work or anything, but I was very curious.)