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I came across this simple proof of Fermat's last theorem. Some think it's legit. Some argued that the author's assumptions are flawed. It's rather lengthy but the first part goes like this:

Let $x,y$ be $2$ positive non-zero coprime integers and $n$ an integer greater than $2$. According to the binomial theorem:$$(x+y)^n=\sum_{k=0}^{n}\binom{n}{k}x^{n-k}{y^k}$$ then,$$(x+y)^n-x^n=nx^{n-1}y+\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}{y^k}+y^{n}$$ $$(x+y)^n-x^n=y(nx^{n-1}+\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}y^{k-1}+y^{n-1})$$

$$y(nx^{n-1}+\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}y^{k-1}+y^{n-1})=z^n$$

In the first case, he assumed that the 2 factors are coprime when $\gcd(y,n)=1$ . Then he wrote: $$y=q^n$$ $$ nx^{n-1}+\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}y^{k-1}+y^{n-1}=p^n$$ By replacing $y$ by $q^n$, \begin{equation} nx^{n-1}+\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}q^{n(k-1)}+q^{n(n-1)}=p^n (*) \end{equation}

from this bivariate polynomial,he fixed alternatively $x$ and $y=q^n$ and by applying the rational root theorem, he obtained $$q^{n(n-1)}-p^n=nxt $$ and

$$ nx^{n-1}-p^n=q^ns $$ ($s,t$ non-zero integers) by equating $p^x$: $$ q^{n(n-1)}-sq^n=nx(t-x^{n-2})$$ Then, he uses one of the trivial solutions of Fermat's equations. He wrote, when $x+y=1$,if $x=0$ then $y=1$ and vice versa.

Therefore, he wrote: $x=0$ iff $q^{n(n-1)}=sq^n$, he obtains: $$q=1$$ or $$s=q^{n-2}$$

By substituting $s$ by $q^{n-2}$ in $nx^{n-1}-p^n=q^ns$, he obtains: $$nx^{n-1}-p^n=q^{n(n-1)}$$ Then, he replace that expression in equation (*) and pointed out that:$$\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}q^{n(k-1)}=0$$. Since $x,y=q^n$ are positive integers for all $n>2$, a sum of positive numbers can not be equal to zero. Which leads to a contradiction.

What do you think?

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    $\begingroup$ Where does $y(nx^{n-1}+\sum_{k=2}^{n-1}\binom{n}{k}x^{n-k}y^{k-1}+y^{n-1})=z^n$ come from? $\endgroup$ – Mathmo123 Feb 8 '15 at 16:26
  • $\begingroup$ Just skimming this, it looks like it never actually uses the hypothesis that $n>2$, which would mean it's invalid. It just tosses "and $n>2$" on at the end, after it, apparently, proves that $x^n+y^n=z^n$ has no solutions for all $n$. $\endgroup$ – Milo Brandt Feb 8 '15 at 16:29
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    $\begingroup$ Do you have any reference where this comes from? And what have you done about it? Which steps can you reproduce and where do you have questions or doubts? $\endgroup$ – Martin R Feb 8 '15 at 16:44
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    $\begingroup$ we now $(x+y)^n-x^n=z^n$contradiction proves $x=0 iff y=1$.but it is trivial.whats the relation of this to last fermat theorem $\endgroup$ – ali Feb 8 '15 at 16:53
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    $\begingroup$ "In the first case, he assumed that..." I see no second case. $\endgroup$ – RghtHndSd Feb 8 '15 at 16:58
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Notice that the proof is actually looking for solutions to $(x+y)^n-x^n=z^n$ which is equivalent (though it would be nice if the proof writer had, I don't know, stated this) - you can see this when it goes from $$(x+y)^n-x^n=\text{stuff}$$ to $$z^n=\text{the same stuff}$$ in the fourth equation.

The first serious error in the proof is assuming that $\gcd(y,n)=1$. It gives no justification for this and it does not appear clear to me that proving the theorem in this case implies the general theorem.

The next error is somewhat more serious - he takes two equations from the rational roots theorem which are presumably correct under the assumption that $\gcd(y,n)=1$, and then considers only a single solution of them, rather than the general solution. He takes $x+y=1$ at this points, and all his further work relies on that assumption. So now, we are proving the following statement: $$1^n-x^n=z^n$$ has no solutions in the positive integers.

That's not Fermat's Theorem, and I think most any reader can come up with a much shorter proof of the fact. Notably, upon close examination of the proof, it does never use the hypothesis that $n>2$, and hence must be false. Yes, even $1^2-x^2=z^2$ has no solutions in the positive integers, and the proof tries to conclude from there as $(x+y)-x^2=z^2$ has no solutions in the positive integers - but, oh wait...

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    $\begingroup$ @Ramunjndscpl That's not really true; it's the empty sum, which equals zero. Notice that the author gets there by subtracting out all the terms of a non-empty sum - so it's perfectly natural that that term would be $0$ - and that's exactly what the proof concludes. The author's logic does not rely on that sum being non-empty, hence the conclusion goes through for $n=2$ too. $\endgroup$ – Milo Brandt Feb 8 '15 at 17:29
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There is a "trick", due to Marc Krasner, which prevents you from wasting time in examining "elementary" arithmetic proofs of Fermat's Last Theorem. "Elementary" means precisely that the proof uses only addition and multiplication (operations in a ring), and perhaps also the existence and unicity of decomposition into prime factors (so the ring in question is factorial). I suppose this is the case here, although not all details are given. Then, without checking anything, you can be assured that the reasoning is certainly wrong. This is because all such "elementary" arguments can be repeated word for word in the ring $Z_p$ of p-adic integers, which is factorial (and a lot more !), but in which FLT is false, because in the field $Q_p$ of p-adic numbers, the equation $x^p + y^p = 1$ always has non trivial solutions (if you take $y$ to be a high power of $p$, then p-adic analysis tells you that $1 - y^p$ has a p-th root in $Q_p$).

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