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Some elementary manipulation of Fermat's theorem is tried and an approach to prove it in very few lines is suggested. The question is to pinpoint errors in the analysis, as usually these stuff is junk. Anyhow algebra is fun and I hope you'll enjoy the faulty ride. Can you see the mistake? I can't. I verified the last explicit formula for $n=3$ and $a=7$, $b=13$, $c = (a^3+b^3)^{1/3}$
Here is the approach:
Consider the following equation over integers with $a \in \mathbb N, b \in \mathbb N, a < b, c \in \mathbb N, n \geqslant 2, \text{odd}, \gcd(a,b)=1, \gcd(a,c)=1,\gcd(b,c)=1$, $$ a^n + b^n = c^n. $$ Put $y = b / c$, $r = a / b$ and we get $$ y^k = \frac{1}{1 + r^k} = \frac{1}{(1 + r) (1 - r + r^2 -r^3+ \cdots + r^{k - 1})} = \frac{1}{(1 + r) u_{k - 1} (r)}. $$ Now we can rewrite this as $$ y^k u_{k - 1} (r) = \frac 1{1+r} = \frac {1+r^k - r^k}{1+r} = u_{k - 1} (r) - r^k \frac{1}{1 + r} = u_{k - 1} (r) - r^k b / (b + a). $$ Multiply by $c^k b^{k - 1}(b+a)$ yields, $$ (b + a) b^k b^{k - 1} u_{k - 1} (r) = (b + a) c^k b^{k - 1} u_{k - 1} (r) - c^k a^k. $$ Now note that, $$ b^{k - 1} u_{k - 1} = b^{k - 1} - b^{k - 2} a^1 + \ldots \dotplus a^{k - 1} = b v (a, b) + a^{k - 1}, $$ And using this we conclude, $$ (b + a) b^k (b v + a^{k - 1}) = (b + a) c^k b v + (b + a) c^k a^{k - 1} - c^k a^{k} = (b + a) c^k b v + c^k b a^{k - 1}. $$ Divide by $b$, $$ (b + a) b^{k - 1} (b v + a^{k - 1}) = ((b + a) v + a^{k - 1}) c^k $$ Now, if $p \backslash (a + b)$ Then either $p \backslash c$ or $p \backslash a$. If the latter then $p \backslash b$ which contradicts $\gcd (a, b) = 1$.
Hence we can write $$ c = \alpha (a + b) $$ But $c < a + b$ hence a contradiction. This shows that Fermat's last theorem holds for odd $n$ larger then 2. Then it holds also for every $n$ that has an odd prime as a factor and the remaining cases follows from the special case $n = 4$ and Fermat's last theorem is proven.
I am not sure if this is the crux of the problem, but in the final step the implication
$$(p \mid (a+b) \implies p\mid c)\implies c = \alpha(a+b)$$
is false in general if $\alpha$ is expected to be an integer.
For example, take $a+b = 27$ and $c=15$. (If you insist, take $a = 13, b=14$) then every prime that divides $a+b$ also divides $c$, but $c$ is not a multiple of $a+b$; moreover $a+b>c$ which does not lead to a contradiction.