Flt

[kkris1]

I have a proposition to prove FLT

See attachement

Fermat_Theorem1.doc

[kkris1]

Some people pointed out there could be some ambiguity in my statements. So I decided to revise it. Could you please help me to find any errors in it?

Fermat Theorem1.doc

[Klaynos]

We can see immediately there are no integer numbers for n>2 to satisfy the equation.

 

Show me mathematically how you can see this...

[kkris1]

d^2(X^(n-2)+Y^(n-2))=Z^n

but

Z^n=X^n+Y^n

so

d^2(X^(n-2)+d^2(Y^(n-2))=X^n+Y^n

which is clearly impossible if X does not equal Y

[the tree]

[math]d^{{2x^{n-2}}+y^{n-2}}=z^n[/math]

but

[math]z^n=x^{n}+y^n[/math]

so

[math]d^{{2x^{n-2}}+y^{n-2}}=x^{n}+y^n[/math]

which is clearly impossible if x does not equal y

You'd be saving yourself a lot of effort and making everything clearer if you use the LaTeX system.

I don't see how this is "clearly" impossible.

[Klaynos]

d^2(X^(n-2)+Y^(n-2))=Z^n

but

Z^n=X^n+Y^n

so

d^2(X^(n-2)+d^2(Y^(n-2))=X^n+Y^n

which is clearly impossible if X does not equal Y

 

Can you please show mathmatically how that is clearly impossible?

 

incidently for quick super and subscript the [ sub ] and [ sup ] tags are very usefull...

[kkris1]

See attachement . It should clarify the issue

Fermat_Theorem1.doc

[cosine]

d^2(X^(n-2)+Y^(n-2))=Z^n

but

Z^n=X^n+Y^n

so

d^2(X^(n-2)+d^2(Y^(n-2))=X^n+Y^n

which is clearly impossible if X does not equal Y

The thing is when you already assume that x=y when you say "Consider the point P(x' date=' y) where [b']x=y[/b]=d"

 

So of course it only holds when x=y, you assumed it so!

 

Therefore your "proof" does not prove anything.