Fermat's last theorem as per the wiki states that
No three positive integers a, b, and c satisfy the equation $a^n + b^n = c^n$ for any integer value of $n>2$.
I just recently came across the history of this and like many before me, I had a go at it. I am quite sure, that I have made mistakes. Can someone help me figure out holes in my assumptions. Thanks.
Concretely, can we reject cases where $a+b<c$ and $a+b=c$ that the above scenario will not satisfy. This I did using the binomial expansion of degree $n$ which hypothetically satisfies the equation. However, LHS left with the non-zero "other" terms which creates a contradiction.
The solutions can thus potentially hold for the $a+b>c$ case. Now, we show that the only case where the solutions can further exist are if $a,b,c$ are sides of a triangle. For this, we show that triangle in-equality must hold for all combinations of $\{a,b,c\}$. Effectively, $a+b>c, b+c>a, a+c>b$. This is because if $b+c<a$, then $(b+c)^n < a^n$ which implies $2b^n+\gamma < 0$ where $\gamma$ is the positive term from binomial expansion. This is not possible with the initial assumptions we have for $a,b,c,n$. Therefore, $a,b,c$ must be sides of triangles holding triangle in-equality. Furthermore, $a < c$ and $b < c$. Otherwise, $a^n>c^n$ or $b^n>c^n$ which supports the theorem. These two inequalities imply $c$ is the largest side.
We again start with the assumption that we can find solutions to Fermat's equations. For a right-angled triangle. $$a^n-a^2c^{n-2}+b^n-b^2c^{n-2}=0$$ $$\implies a^2(a^{n-2}-c^{n-2})+b^2(b^{n-2}-c^{n-2}) = 0$$
Two negatives in the parenthesis cannot sum up to 0.
Last case left is of the obtuse triangle. Let the height be $x$ and "total base" be $b+y$. We have $x^2+y^2=a^2$ and $x^2+(b+y)^2 = c^2$. With these two equations we arrive at
$$ a^n - a^2c^{n-2}+b^n-b^2c^{n-2} = 2bc^{n-2}y$$ $$ \frac{a^n-a^2c^{n-2}b^n-b^2c^{n-2}}{2bc^{n-2}} > 0$$ Since, $y > 0$ as we are dealing with an obtuse triangle.
$$ a^n - a^2c^{n-2}+b^n-b^2c^{n-2} > 0$$ Again this inequality cannot be satisfied under our starting conditions.
Lastly, the case of acute triangle, we use the inequality $a^2+b^2>c^2$ for an acute triangle. $a^{n-2}a^2 + b^{n-2}b^2 = c^{n-2}c^2 > c^{n-2}(a^2+b^2)$
$$\implies a^2(a^{n-2}-c^{n-2})+b^2(b^{n-2}-c^{n-2}) > 0$$ which is not possible since $c$ is the largest side.
I am not sure which conditions I am missing or for the ones considered what wrong assumption did I make.