kkris1 Posted April 11, 2006 Share Posted April 11, 2006 I have a proposition to prove FLT See attachement Fermat_Theorem1.doc Link to comment Share on other sites More sharing options...
kkris1 Posted April 13, 2006 Author Share Posted April 13, 2006 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 Link to comment Share on other sites More sharing options...
Klaynos Posted April 13, 2006 Share Posted April 13, 2006 We can see immediately there are no integer numbers for n>2 to satisfy the equation. Show me mathematically how you can see this... Link to comment Share on other sites More sharing options...
kkris1 Posted April 13, 2006 Author Share Posted April 13, 2006 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 Link to comment Share on other sites More sharing options...
the tree Posted April 14, 2006 Share Posted April 14, 2006 [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. Link to comment Share on other sites More sharing options...
Klaynos Posted April 14, 2006 Share Posted April 14, 2006 d^2(X^(n-2)+Y^(n-2))=Z^nbut 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... Link to comment Share on other sites More sharing options...
kkris1 Posted April 18, 2006 Author Share Posted April 18, 2006 See attachement . It should clarify the issue Fermat_Theorem1.doc Link to comment Share on other sites More sharing options...
cosine Posted May 4, 2006 Share Posted May 4, 2006 d^2(X^(n-2)+Y^(n-2))=Z^nbut 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. Link to comment Share on other sites More sharing options...
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