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One of my dreams in life has been to see a proof of Fermat's Last Theorem that doesn't require fancy and advanced mathematics. After years of trying off and on (mostly off), I haven't been able to come up with one, so I decided to search the internet for one. After doing such, I found the following paper, which appears to give a reasonable approach. And it's written very well too. http://www.relativitydomains.com/Mathematics/Fermat/Fermat.pdf

Can anyone find any mistakes in the proof?

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I didn't look at the paper, but yes, I'm sure there are many mistakes. –  Graphth Oct 5 '12 at 15:46
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To show that a certain method works for small $n$ and then just say that it extrapolates to higher $n$, is not a proof. The same method could beused to show that every polynomial can be solved by radicals, because it is possible for degree 1, 2, 3, and 4. –  Hagen von Eitzen Oct 5 '12 at 15:47
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The justification of $(2.1)\Rightarrow(2.2)$ is marvellous! –  Hagen von Eitzen Oct 5 '12 at 15:48
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Craig, just so you understand, many believe that Fermat actually did know a proof for the small case $n=4$, using the method of descent. And, by the way, this method isn't all that complicated. So, for someone else to come up with a "proof" that only works in a small case and claim it works for all, it's not a step in the right direction at all. It's nothing new. –  Graphth Oct 5 '12 at 15:56
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It's arguably naive to believe a three-year old paper claiming an elementary proof of FLT (with other alarm bells) could plausibly be sound; a question I think would be better received would be something like "what is the error in this attempt"? Unfortunately, the downvoters are going to come and punish you merely for naivete... –  anon Oct 5 '12 at 15:57
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2 Answers

up vote 6 down vote accepted

Here's one, I think:

after (2.19) the author writes $$a = (x + \frac{b}{2})(1 - \frac{1}{q_3}) + \frac{b}{2}$$ where $b$ is an odd integer, $x$ is an integer and $q_3 > 1$, and claims that $a$ cannot be an integer. This is clearly wrong: for example, if $x = 1$, $b = 1$ and $q_3 = \frac{3}{2}$, it follows $a = 1$.

The rest of the paper seems to be based on this assertion..

Maybe there's some additional conditions that I don't see, though.

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The author does claim that $q_3$ is a 'pure number' but I don't think he assumes it to be an integer –  Cocopuffs Oct 5 '12 at 15:54
    
That's good. Maybe this problem is fixable, maybe it's not. My bet is it's not. –  Craig Feinstein Oct 5 '12 at 16:58
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The author contradicts himself when writing

Application of Descartes’ Rule of Signs shows that (2.3) contains just one positive root, with all the rest being negative. Also, from the co-efficient of $x^{n−1}$ it is clear that all the roots must contain the unique term $(a − b)$. Furthermore, the nature of the other co-efficients also shows that the remaining roots cannot all be real.

If not all roots are real, you cannot say that one root is positive and all other roots are negative.

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