Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

$$x^n=z^n-y^n=(z-y)(z^{n-1}+yz^{n-2}+\cdots+y^{n-1})\ge ny^{n-1}\gt xx^{n-1}$$ contradiction.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.