Where is the problem in this proposed elementary proof of Fermat's Last Theorem [closed]

I stumbled across a website by a chap called Tom Ballard in which he presents his proof of FLT based on elementary techniques: http://www.fermatproof.com

The style is rather 'non-standard', shall we say, and makes it difficult to assess. I have checked through it and have a couple of points to investigate further, but certainly the first part on pythagorean triples is interesting, and correct.

Has anybody else seen it and put in some effort to see if it is correct?

• What exactly are you asking for? – Rasmus Feb 8 '12 at 14:31
• I have not really read any details, but the fact that he mentions that his approach to pythagorean triplets is new, which has been confirmed by many "math people", and the fact that he at some point goes into detail of what a "reduction ad absurdum" proof involves, makes me very skeptical. – Tobias Kildetoft Feb 8 '12 at 14:38
• I do get, on average, about two purpoted elementary proofs of FLT in the mail every year. All of them start with considerations about Pythagorean triplets (none of them seems to be aware of the fact that conics are rational curves, though) and go very wrong right away. This one, at least, has some nice pictures. – Andrea Mori Feb 8 '12 at 15:40
• Let's be clear, I do understand that the writing style of the author is very bad (I have a PhD and understand how to write good mathematics). But I have put in some effort to struggle through his reasoning and have not so far found any major problem in his approach. – Robert Feb 8 '12 at 16:34
• Voting to close. Please see this: meta.math.stackexchange.com/questions/2290/…. If you have any specific mathematical points you want to discuss, please post that. A blanket "is it correct?" type of question is liable to be closed. – Aryabhata Feb 8 '12 at 16:36

The problem is that (4) is used but not proved. (1) through (3) are merely three different versions of a definition of $r$: (1) and (2) are rearrangements of each other, and (3) is obtained from (1) or (2) by adding $z-x$ or $z-y$ to (1) or (2), respectively. So the only "Pythagorean" content is in (4). While a lot of effort is expended on showing that (1) to (3) obtain in the cubic case (which is unsurprising since they merely express the definition of $r$), (4) is just pulled out of thin air and used to claim that $x,y,z$ have to form a Pythagorean triple, when in fact (4) can only be derived by assuming that they do.
• @Robert The flaw to me looks like the author is presenting a specific parametrization of the Pythagorean triples (it looks like just a variant on the usual $(m^2-n^2, 2mn, m^2+n^2)$ to me, with the 'root' $r$ looking like $m^2-2mn$ at first glance though I haven't really gone through the arithmetic closely) and then saying that this parametrization doesn't extend to the higher-powers case - but there's no reason to believe that any parametrization would have to look similar to that, and ample reason not to. – Steven Stadnicki Feb 8 '12 at 17:43
• @Robert: He isn't projecting the case for general $n$ onto the case for $n=2$. He's projecting it into two dimensions and drawing figures that look like the ones he drew for $n=2$, but the relationship $x^2+y^2=z^2$ has no basis in this case, and the corresponding relationship involving $r$ is just introduced without any justification. – joriki Feb 8 '12 at 20:51