Let it be $a$, $b$ and $c$ some positive integers such that $a$, $b$ and $c$ are pairwise coprime, and such that
$$a^{n}+b^{n}=c^{n}$$
Let us assume that $a^{n}$ and $c^{n}$ are the odd numbers of the equation. Therefore, we can express that $b=2^{x}I$, where $2^{x}$ is the higher power of two dividing $b$ and $I$ is the remaining odd integer dividing $b$, and thus $b^{n}=2^{nx}I^{n}$.
Operating, we can state that
$$b^{n}=c^{n}-a^{n}$$
It can be seen that $2^{nx}$ must divide exactly the expression $c^{n}-a^{n}$.
By the binomial expansion, it can be noted that $c^{n}-a^{n}$ for $n\geq3$, $n$ odd, can be factored as $\left(c-a\right)\left(c^{n-1}+c^{n-2}a+...+ca^{n-2}+a^{n-1}\right)$, where the numbers of terms in the second parenthesis is odd; and thus, the sum inside the second parenthesis is odd.
As a result, the higher power of two dividing $c^{n}-a^{n}$ is the same as the higher power of two dividing $c-a$.
Therefore, it follows that necessarily $2^{nx}$ must divide exactly $c-a$.
It can be easily proved that, if $a^{n}+b^{n}=c^{n}$, then $c<a+b<2c$. Therefore, $c-a<b$.
As a result, we get that the equation $a^{n}+b^{n}=c^{n}$ can have solutions for $n$ being some odd prime number only for values of $b>2^{nx}$ (and thus, $c>2^{nx}$), $b$ having some odd factor $I>2^{n-1}$.
Is this lower bound rightly proved? Thanks!