Fermat's Last Theorem states that there exist no positive integers $x, y, z$ such that $x^n + y^n = z^n$, for integer $n > 2$.
If we allow for $x, y, z \in \mathbb{R}_{>0}$, it is obvious that there are infinitely many solutions (take arbitrary $x, y$ and choose $z = \sqrt[n]{x^n + y^n}$).
However, do there always exist solutions with arbitrarily near-integer values? In other words, for each integer $n>2$ and real number $\varepsilon > 0$, we want to find $x, y, z \in \mathbb{R_{>0}}$ so that:
- $|x-x'| < \varepsilon, |y-y'| < \varepsilon, |z-z'| < \varepsilon$ with $ x', y', z' \in \mathbb{Z}_{>0}$
- $x^n + y^n = z^n$
Best Answer
I guess its true. Let me try a proof: Given $n\in\mathbb{N}$ with $n\geq 3$ and given $\varepsilon>0$. Pick $x\in\mathbb{N}$ and $y=1$. Then set $$z=(x^n+1)^\frac{1}{n}.$$ If we show $$\lim_{x\rightarrow\infty,x\in\mathbb{N}}|(x^n+1)^\frac{1}{n}-x|=0.$$ we can choose $x$ to be big enough, such that $$|z-x|=|(x^n+1)^\frac{1}{n}-(x^n)^\frac{1}{n}|<\varepsilon.$$ Since $x\in\mathbb{N}$ the result follows. Let us take a look at the limit: The mean value theorem applied to $t\mapsto t^\frac{1}{n}$ yields a $\xi\in[x^n,x^n+1]$ such that $$|(x^n+1)^\frac{1}{n}-(x^n)^\frac{1}{n}| = \frac{1}{n}(\xi)^{\frac{1}{n}-1}(x^n+1-x^n).$$ Since the exponent $\frac{1}{n}-1$ is negativ, we get $$|(x^n+1)^\frac{1}{n}-(x^n)^\frac{1}{n}|\leq\frac{1}{n}(x^n)^{\frac{1}{n}-1}$$ Again with $\frac{1}{n}-1$ being negative, the result follows.