So, about Fermat's Last Theorem where a,b,c are integers
\({ a }^{ n }+{ b }^{ n }={ c }^{ n }\quad Let\quad x+1=a,\quad y+1=b,\quad z=c,\) so \(a,b>1\)
We have
\({ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }=0\)
Consider
\({ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }=\quad { \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left( x+1 \right) { \left( y+1 \right)}^{ n-1 }\\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 2 \end{matrix} \right) { \left( x+1 \right)}^{ 2 }{ \left( y+1 \right)}^{ n-1 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +...\\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ n-1 \end{matrix} \right) { \left( x+1 \right)}^{ n-1 }{ \left( y+1 \right)}^{ 1 }\)
then
\({ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }>{\left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }\)
Also,
\({ \left( x+1 \right)}^{ n }={ x }^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { x }^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) x+1\\ { \left( y+1 \right)}^{ n }=y^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) y^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { y }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) y+1\\ { \left( z+1 \right) }^{ n }=z^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) z^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { z }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) z+1\)
We have,
\({ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }=\quad { x }^{ n }+y^{ n }-{ z }^{ n }\\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left[ { x }^{ n-1 }+\left( x+1 \right) { \left( y+1 \right)}^{ n-1 }+{ y }^{ n-1 } \right] \\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 2 \end{matrix} \right) \left[ { x }^{ n-2 }+{ \left( x+1 \right)}^{ 2 }{ \left( y+1 \right)}^{ n-2 }+{ y }^{ n-2 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +...\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ n-1 \end{matrix} \right) \left[ x+{ \left( x+1 \right)}^{ n-1 }\left( y+1 \right) +y \right] \\\quad \quad \quad \quad \quad \quad \quad \quad \quad +2\)
\({ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }=\quad { x }^{ n }+y^{ n }-{ z }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left[ { x }^{ n-1 }+{ f }_{ 1 }\left( x,y \right) +{ y }^{ n-1 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 2 \end{matrix} \right) \left[ { x }^{ n-2 }+{ f }_{ 2 }\left( x,y \right) +{ y }^{ n-2 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +...\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ n-1 \end{matrix} \right) \left[ x+{ f }_{ n-1 }\left( x,y \right) +y \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)} \)
then
\({ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }>{ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }\)
\({ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }>{ x }^{ n }+{ y }^{ n }-{ z }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)} \)-----------(2a)
\({ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }>{ x }^{ n }+{ y }^{ n }-{ z }^{ n }\) ---------------------------(1)
\({ x }^{ n }+y^{ n }-{ z }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}>{ x }^{ n }+y^{ n }-{ \left( z+1 \right) }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)} \) -----------(2b)
and so, when
\({ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }=0\) -------------(S)
then from expression (2a) and (2b)
\({ x }^{ n }+y^{ n }-{ \left( z+1 \right) }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}<0\)
and from expression (1)
\({ x }^{ n }+{ y }^{ n }-{ z }^{ n }<0\)
that is \({ \left( z+1 \right)}^{ n }-{ z }^{ n }<2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}\)
\({ \left( z+1 \right)}^{ n }-{ z }^{ n }=\left( \begin{matrix} n \\ 1 \end{matrix} \right) z^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { z }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) z+1\)
\(=\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)z^{ n-i }+1}<2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}\) -------------(3)
for \(n=2\)
\(\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) z+1<\sum _{ 1 }^{ 1 }{ \left( \begin{matrix} 2 \\ i \end{matrix} \right)} \quad 2z<3\) which can be true. (This is a statment that was derived from the starting condition (S) that prove to be possible.)
but for \(n=3\)
\(\left( \begin{matrix} 3 \\ 1 \end{matrix} \right) z^{ 2 }+\left( \begin{matrix} 3 \\ 2 \end{matrix} \right) z+1<2+\sum _{ i=1 }^{ 2 }{ \left( \begin{matrix} 3 \\ i \end{matrix} \right)} \)
\( 3{ z }^{ 2 }+3z\nless7\) for all \(z > 1\).
In fact we have a contradiction for all \(n>2\) because the expression \(\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)} \) is just the coefficients of the \(z\) on the L.H.S. of expression (3). (For \(n>2\) the derived statment proved to be false, invalidating the starting condition (S).) And so, \({ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }\neq{ z }^{ n }\) for \(n>2\).
Therefore,
\({ a }^{ n }+{ b }^{ n }={ c }^{ n }\) has no valid solution for \(n>2\) and \(a,b>1\)
And that's Fermat's Last Theorem, proved by contradiction.
(Expression (3) is derived from (S), it is not an attempt to solve for \(z\).)
