Fermat's Last Theorem solved through induction on $z$, not on $n$

Question

Assume that we have shown a proof for the case $z=1$ and $z=2$ for $x^n +y^n = z^n$.

Can we say that via mathematical induction on $z$ that this is now true for all integers $z$ for $n>2?$

Answer 1

I'm afraid this isn't quite how proof by induction works.

Call $P$ whatever we're proving about $z$. We can't simply say "$P$ is true for $z=1$ and $z=2$, therefore it's true for all $z>2$". (For example, "$z<3$" is true for both of them but false for $z>2$.)

To prove $P$ by induction we have to show that $P$ being true for one case forces it to be true for the next. Then we say 'Well it's true for $z=1$ so it must be true for $z=2$, so it must be true for $z=3$, so . . . " and let this count its way through all the remaining positive integers. But we've not got infinitely long in which to do the counting, so we compress that into " so by induction, it's true for all $z\geq 1$".

The key point is that each case must prove the next one, so there's a chain all the way up from a case that we already know to be true.

So I'm afraid your proof needs to do rather more.

Answer 2

Mathematical induction isn't "It works for $1$, it works for $2$, so it always works." You have to prove that the statement in question being true for $k$ implies that it is also true for $k+1$ - for any value of $k$ (or an equivalent variant of this).

What you seem to be doing is assuming that the induction step is given as proved already, which isn't the case - usually in fact this is precisely the nontrivial part of a proof by induction.

So no, you cannot say anything about FLT.