Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
8  
Does FLT for polynomials state that $x(t)^n + y(t)^n \not= z(t)^n$ for all $t$? – Brian Tung Jan 19 at 22:43
    
All nonzero $t$, i guess. It may be what i'm missing. – User1 Jan 19 at 22:44
    
Minor nitpick, you should say "entail" rather than "imply". Since FLT is true for integers, anything implies it. – DanielV Jan 19 at 22:46
4  
@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. – Brian Tung Jan 19 at 22:53
    
@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 $a$”. – egreg Jan 19 at 22:55
up vote 11 down vote accepted

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.