# Diophantine Equation $x^n+y^n=z^n$

Problem

Using simple mathematical operators (+,- ,> etc.) can it be shown that (assuming $x<y$) Fermat’s theorem is always true when $$n\ge x$$

Request I am sure this approach has been discussed somewhere earlier. If anyone therefore could either direct me to such resource of discussion or show the proof. I have also developed a proof which I shall share tomorrow for a review.

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$$x^n=z^n-y^n=(z-y)(z^{n-1}+yz^{n-2}+\cdots+y^{n-1})\ge ny^{n-1}\gt xx^{n-1}$$ contradiction.
Thanks Gerry, my answer is also along identical line. I have used the inequality $$\frac{x}{n} +y \gt z$$ This shows that when $x <n$ then $$\frac{x}{n} < 1$$ $$\Rightarrow 1+y>\frac{x}{n}+y>z>y$$ Showing that x lies between $y$ and $y+1$ which is a fallacy considering $z$ is an integer. This brings me to the next pertinent question. What happens when $n < x$. Is there any discussion on this anywhere ? – Barun Dasgupta Aug 12 '12 at 13:29
You're asking whether there's any discussion of Fermat's equation when $n\lt x$? Considering that we've done the $n\ge x$ case by school algebra, I'd say every serious discussion of Fermat, including the work of Wiles, is a discussion of what happens when $n\lt x$. But perhaps I misunderstand your question. – Gerry Myerson Aug 12 '12 at 13:38