[Math] Why doesn’t Fermat’s Last Theorem for polynomials entail that for integers

abstract-algebranumber theory

Fermat's Last Theorem for polynomials follows from the Stothers-Mason theorem, that is: For any integer $n\geq 3$, there do not exist polynomials $x(t), y(t), z(t)$ not all constant such that $x(t)^n + y(t)^n = z(t)^n$ for all $t\neq 0$.

But since we can always find suitable polynomials in $t$ such that $(x(t), y(t), z(t)) = (a,b,c)$, why can't FLT for polynomials entail that for integers ?

For example, suppose we had $7^n + 8^n = 15^n$ for some integer $n\geq 3$, we would have $t=2$ such that $(7, 8, 15) = (3t+1, 4t, 8t-1)$, which is impossible by FLT for polynomials ?

Best Answer

The Mason-Stothers theorem holds for polynomials over $\mathbb{Q}$, and any $n\ge3$. Note that $\mathbb{Q}$ is an infinite field, so two polynomials $P(t)$ and $Q(t)$ are different if and only if there is $a\in\mathbb{Q}$ such that $P(a)\ne Q(a)$.

So FLT for polynomials over $\mathbb{Q}$ can be stated

for any coprime and nonconstant polynomials $x(t)$, $y(t)$, $z(t)$ over $\mathbb{Q}$ and any $n\ge 3$, there exists $a\in\mathbb{Q}$ such that $$x(a)^n+y(a)^n\ne z(a)^n$$

because this is an alternative way for saying $x(t)^n+y(t)^n\ne z(t)^n$ as polynomials.

So the theorem does not say that the inequality holds for all $a\in\mathbb{Q}$, which would be needed to derive FLT from it.

Another way for looking at it is considering FLT for polynomials over $\mathbb{R}$: can you perhaps derive from it that for $a,b,c\in\mathbb{R}$ and $n\ge 3$ we have $a^n+b^n\ne c^n$?