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?
Best Answer
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 $Q_p$).