Can someone explain to me why this does not constitute a proof of Fermat's Last Theorem, please?

Basically, using something I've read online, it appears you can write an equation for $(a, b, c)$ to find solutions for equations $a^n + b^n = c^n$ in the form of $$a = (u^2 - v^2)^{2/n}, \: b = (2uv)^{2/n},\: c = (u^2 + v^2)^{2/n}$$

which won't produce integer solutions for $n > 2$.

Could someone explain to me the mistake in this "proof" please?

The full work is here (there's no downloading required because it's just a PDF).

Thanks in advance

  • 10
    $\begingroup$ Without looking at the pdf, what you've suggested wouldn't count as a proof because all it verifies is that that construction doesn't give integer solutions for $n>2$. But that doesn't guarantee no other construction could work. (None does, in fact, but proving that is the hard part!) $\endgroup$ Jun 15, 2016 at 17:15
  • 4
    $\begingroup$ Showing that one particular method of producing solutions doesn't work does not show that there's no solution. So your proof is just no proof at all. But I disagree with the downvote. This is exactly the sort of thing that people do get confused about - having an explanation of what the error is available seems like a good thing. $\endgroup$ Jun 15, 2016 at 17:15
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    $\begingroup$ Another way of looking at the failure in the proof (which is complete and unrecoverable) is that the algebraic manipulations may be valid but at no point does it establish that $u,v$ must be integers. Thus it is very far from obvious that $(u^2-v^2)^{2/n}$, etc. cannot produce an integer. $\endgroup$
    – Erick Wong
    Jun 15, 2016 at 17:22
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    $\begingroup$ @dolphinsupreme $u,v$ are derived from $k$. In turn, $k$ is defined as the ratio between $y^{n/2}$ and $1 + x^{n/2}$. Notice all the square roots in those expressions? There's absolutely no reason to believe $k$ is rational. $\endgroup$
    – Erick Wong
    Jun 15, 2016 at 17:31
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    $\begingroup$ The final justification for why $(2uv)^{2/n}$ cannot be an integer is also horribly, horribly incomplete and unconvincing. By the "logic" of that paragraph, one might as well replace the entire proof by an appeal to "we can see this trend". $\endgroup$
    – Erick Wong
    Jun 15, 2016 at 17:33

1 Answer 1


The argument seems to implicitly assume that $u$ and $v$ are integers. This assumption is not justified, as explained below. But even accepting this assumption, there is a fatal flaw in the final step of the argument. The argument given is that $b$ cannot be an integer if $n>2$, because of the irrational factor $2^{2/n}$ appearing in the expression $b=(2uv)^{2/n}$. However, this factor might combine with irrational parts of $u^{2/n}$ and $v^{2/n}$ to produce an integer. For instance, take $n=3$, $u=4$, and $v=1$.

Moreover, the argument never actually defines exactly what $u$ and $v$ are. Instead, it defines $$k=\frac{y^{n/2}}{1+x^{n/2}}$$ and later writes $k=\frac{u}{v}$. So in order to be able to choose $u$ and $v$ to be integers, you would need to know $k$ is rational. From the definition of $k$, there is no reason to expect it to be rational if $n$ is odd, since its definition involves taking square roots of $x$ and $y$. There is no problem if $n$ is even, but the issue raised in the previous paragraph is still a problem then.

  • $\begingroup$ Thank you. This is exactly what I was looking for. Very explicit, and I like the example. $\endgroup$ Jun 15, 2016 at 17:43

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