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 ?

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    $\begingroup$ Does FLT for polynomials state that $x(t)^n + y(t)^n \not= z(t)^n$ for all $t$? $\endgroup$
    – Brian Tung
    Jan 19, 2016 at 22:43
  • $\begingroup$ All nonzero $t$, i guess. It may be what i'm missing. $\endgroup$
    – User1
    Jan 19, 2016 at 22:44
  • $\begingroup$ Minor nitpick, you should say "entail" rather than "imply". Since FLT is true for integers, anything implies it. $\endgroup$
    – DanielV
    Jan 19, 2016 at 22:46
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    $\begingroup$ @User1: No, I mean, there is a difference between "There do not exist polynomials $x(t), y(t), z(t)$ and integer $n \geq 3$ such that $x(t)^n+y(t)^n = z(t)^n$ for all $t \not= 0$" and "For all polynomials $x(t), y(t), z(t)$ and integer $n \geq 3$, $x(t)^n+y(t)^n \not= z(t)^n$ for all $t \not= 0$." I suggest that FLT for polynomials is the former and not the latter. $\endgroup$
    – Brian Tung
    Jan 19, 2016 at 22:53
  • $\begingroup$ @User1 FLT for polynomials tells you that, given coprime nonconstant $x(t)$, $y(t)$, $z(t)$ and $n\ge3$, there exists $a$ such that $x(a)^n+y(a)^n\ne z(a)^n$ (so $x(t)^n+y(t)^n\ne z(t)^n$ as polynomials). Not that “for all $a$”. $\endgroup$
    – egreg
    Jan 19, 2016 at 22:55

1 Answer 1


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 $a^n+b^n\ne c^n$?


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