S. Ramanujan explored how to add two different cubes in two different ways, which made many, including me, think that he was working on FLT. But like always we need to see a direct link, so let's look at FLT for $n=3$. Let's suppose $x^3+y^3=z^3$ has solutions where $x,y,z\in{\mathbb{Z^+}}$. Since $x,y,z\in{\mathbb{Z^+}}$, we can say that $x=a+1$, $y=b-1$, and $z=c+1$ where $a,b,c\in{\mathbb{Z^+}}$ and $b\neq 1$. In particular, \begin{eqnarray*} (a+1)^3+(b-1)^3&=&(c+1)^3\\a^3+3a^2+3a+1+b^3-3b^2+3b-1&=&c^3+3c^2+3c+1\\a^3+b^3&=&c^3+1+3(c^2-a^2+b^2+c-a-b)\\a^3+b^3&=&c^3+1+3k\text{, where }k\in{\mathbb{Z}}. \end{eqnarray*} From this point we see that our solutions, if any, belong to \begin{eqnarray*} a^3+b^3\equiv{c^3+1}\ (\text{ mod 3 )} \end{eqnarray*} which gives us the link between Ramanujan's near misses and FLT. That's not all! By a very wonderful proposition of \begin{eqnarray*} (m+n)^p\equiv{m^p+n^p}\ \text{( mod $p$ ), where $p$ is prime} \end{eqnarray*} and similar thinking from the $n=3$ case we can generalize this idea of near misses for exponents that are odd primes. Carefully observe, \begin{eqnarray*} (a+1)^p+(b-1)^p&=&(c+1)^p\\a^p+pq+1+b^p+pr-1&=&c^p+pv+1\text{, where $p$ is an odd prime and }q,r,v\in{\mathbb{Z}}\\a^p+b^p&=&c^p+1+p(v-q-r)\\a^p+b^p&=&c^p+1+pm\text{, where }m\in{\mathbb{Z}}\\a^p+b^p&\equiv&{c^p+1}\ \text{( mod $p$ )}\ \\(a+b)^p&\equiv&(c+1)^p\ \text{( mod $p$ )} \end{eqnarray*} From this point let's consider a special case where $a+b=c+1$. So, \begin{eqnarray*} (a+1)^p+(b-1)^p&=&(a+1+b-1)^p\\&=&(a+1)^p+(b-1)^p+pw\text{, where }w\in{\mathbb{Z^+}}\\0&=&pw \end{eqnarray*} since $p$ and $w$ are strictly greater than zero, we know that $0=pw$ isn't true. So, $a+b\neq c+1$. Now consider the slightly trickier and a more general case where $l\in \mathbb{Z}$ such that $a+b-pl=c+1$, since $a+b\equiv{ c+1 }\text{( mod $p$ )}$. Consequently, \begin{eqnarray*} (a+1)^p+(b-1)^p&=&(a+1+b-1-pl)^p\\&=&(a+1+b-1)^p-(pl)^p+pf\text{, where $f\in \mathbb{Z}$}\\&=&(a+1)^p+(b-1)^p+pw-(pl)^p+pf\\&=&(a+1)^p+(b-1)^p+p(w-p^{p-1}l^p+f)\\&=&(a+1)^p+(b-1)^p+p\sigma \text{, where }\sigma \in \mathbb{Z}\\0&=&p\sigma \end{eqnarray*} If one could prove that $\sigma\neq 0$, then we could say by similar thinking in the $a+b=c+1$ case, we can clearly deduce that $0\neq p\sigma$. So, we could conclude that $a+b-pl\neq c+1$ where $l\in\mathbb{Z}$. From that point, we could further say since $a+b-pl\neq c+1$ where $l\in\mathbb{Z}$, we know that this contradicts our supposition that FLT isn't true.
My question is does anybody have any insight on how to show that $\sigma\neq 0$ using elementary methods, or show that there something wrong with my reasoning? Any help would be very much appreciated.