Fermat's last theorem states:
(1) $x^n + y^n = z^n$
has no solutions for x, y, z and n positive coprime integers and n > 2. An open question is whether there exists a simple proof hinted at by Fermat. If you can spot an error in my thoughts below, please point it out.
This can be rewritten as:
(2) $x^n+(x+a)^n=(x+b)^n$ with $y=x+a$ and $z=x+b,$ a and b integers, then,
(3) $x^n+(x^n+{n\choose n-1}x^{n-1}a+...+{n\choose r}x^ra^{n-r}+...+a^n)$=
$(x^n+{n\choose n-1}x^{n-1}b+...+{n\choose r}x^rb^{n-r}+...+b^n)$, or rearranging and cancelling $x^n$ on both sides
(4) $x^n-{n\choose n-1}x^{n-1}(b-a)+...-{n\choose r}x^r(b^{n-r}-a^{n-r})+...-b^n+a^n=0$
So if we think about what this would mean for the geometrically tractable case of n=3 for a cube, so substituting n=3 in (4) gives
(5) $x^3-3x^2(b-a)-3x(b^2-a^2))-b^3+a^3=0$
The cube of side x = the sum of the different cuboids between the 2 largest cubes. These differences are composed of descending terms in $x^r$. By taking these terms all on to the LHS, we are saying that the RHS = $0$.
There is another way to create a cube of volume $0$, starting from a cube of side x. If we start with a cube of side x and imagine removing a length of d until x=d, then that will create a volume of $0$. We can write this as:
(6) $(x-d)^3 = 0$, expanding gives
(7) $x^3-3dx^2+3d^2x-d^3=0$
we can then compare coefficients in ${n\choose r}x^r$ between (5) and (7) as these are both expressions describing the same geometric operation of taking $x^3$ and subtracting volumes until RHS = $0$. So, for a given value of d, we have 2 variables a and b constrained by the 3 independent equations:
(8) $d=b-a$
(9) $d^2=a^2-b^2$
(10) $d^3=b^3-a^3$
Therefore this set of equations is over-constrained and has no solutions for a and b and therefore for x, y, z for n=3. With 2 variables it is possible to define 2 of the 3 sides, but then the 3rd is not definable in the integers. Therefore (1) is impossible for n=3.
We can extend this approach for the general case of n. Equations (8), (9) and (10) will always be produced for n>2 as will
$(-1)^rd^r=a^r-b^r$ (general equation)
$(-1)^nd^n=a^n-b^n$ (final equation)
Generally there will be n equations formed when comparing coefficients but always only 2 variables. 2 of the n-dimensional sides are definable, but then the remaining dimensions have no solutions in integers.
The simultaneous equations formed in comparing coefficients between (5) and (7) will always be over-constrained for n>2. Therefore (1) is impossible.
