I just came across an interesting but most certainly wrong elementary proof of Fermat's Last Theorem. After about an hour of scrutinising, i'm yet to discover the mistake (i'm definitely not an expert on this subject). So here it is:
Lemma 1. (Due to Perez-Cacho (see Paulo Ribenboim's `13 Lectures on FLT, pp.83-84). Fermat's Last Theorem for the exponent $2j-1$ where $j \in \mathbb{N}_{\geq 2}$, is equivalent to the statement that the equation $(UV)^{j} = U+V$ has only the trivial solutions in $\mathbb{Q}$, whereby $UV=0$.
Claim 1. Fermat's Last Theorem holds.
Proof. Assume that Fermat's last Theorem is true for every exponent in $[3, m-1)$ (we may take $m\geq 5$ since the exponents 3 and 4 were proved in the 18th century), so that $m$ is supposedly the least positive integer $\geq 3$ such that \begin{equation} W^{m}=U^{m}+ V^m \end{equation} for some $U, V, W \in \mathbb{N}$. Since $m\geq 5$ must be \textit{indivisible} by $12$, note that it must be divisible by some odd prime $2n-1>3$. Hence we infer from $(1)$ that \begin{equation} w^{2n-1}=u^{2n-1}+v^{2n-1} \end{equation} for some positive integers $u, v, w$. Let $x=\frac{wu^{n-1}}{v^n}$ and $y=\frac{wv^{n-1}}{u^n}$, so that \begin{equation} (xy)^n = x+y. \end{equation} Put $n=2b$ where $b \in \mathbb{Q}_{>1}$ and let $d\in [2, 2b)$ be some integer such that $2d-1 \mid 4b-1$, thus $4d-2 \mid 8b-2 = 4(2b-d) +(4d-2)$. Define $c:= \frac{2b-d}{4d-2}$ so that $d=\frac{2(b+c)}{1+4c}\in \mathbb{N}$. Inserting $n=2b$ into the above and multiplying both sides by $(xy)^{2c}$ gives \begin{equation} (xy)^{2(b+c)}=(x/y)^{1+2c}y^{1+4c}+(y/x)^{1+2c}x^{1+4c}. \end{equation} Let $\alpha = (x/y)^{1+2c}y^{1+4c}$ and $\beta = (y/x)^{1+2c}x^{1+4c}$ hence $xy = (\alpha \beta)^{\frac{1}{1+4c}}$. Thus the above becomes \begin{equation} \alpha + \beta=(\alpha \beta)^{\frac{2(b+c)}{1+4c}} = (\alpha \beta)^{d}. \end{equation} By Lemma 1, it follows from the above equation that there should exist some positive integers $X, Y, Z$ such that \begin{equation} X^{2d-1}+Y^{2d-1}=Z^{2d-1}. \end{equation} Recall that $2\leq d<2b=n$ and $d \in \mathbb{N}$. Thus the above contradicts the minimality of $m$ since $3\leq 2d-1<2n-1 \leq m$. This implies that our supposition must be false, so we are done. $\blacksquare$
EDIT: Mistake found, see the comments.