TITLE: Is there a simple way to illustrate that Fermat's Last Theorem is plausible?
QUESTION [7 upvotes]: A first step in proving a theorem is true could be to show that it is plausible, so at least you then would have a general idea that it could be true and have something to start with in proving it. Simply put: if you get the picture, you can do the math.
This brings me to my question:
Is there a simple way to demonstrate that Fermat's Last Theorem is (at least) plausible? And if so, is it thought that Fermat had found it himself? Or has trying to proof FLT always been 'a shot in the dark' for everyone anyway?

REPLY [8 votes]: My understanding is Fermat Last Theorem
$$\not\exists x, y, z \in \mathbb{Z}_{+} : x^p + y^p = z^p\quad\text{ for } p > 2$$ is plausible because integers of $p^{th}$ power thin out too quickly as $x$ getting big (at least for $p > 3$ ).
For any integer $N$, the number of $p^{th}$ power smaller than $N$ is of the order
of $N^{\frac{1}{p}}$. If we randomly pick a integer near an number $N$, the probability that we get a $p^{th}$ power is around $\frac{1}{p} N^{\frac{1}{p}-1}$.
Heuristically, the probability that we get a sum of two $p^{th}$ power should be something like
$$\frac{1}{p^2} \sum_{n=1}^{N-1} n^{\frac{1}{p}-1} (N-n)^{\frac{1}{p}-1}
\sim \frac{1}{p^2} N^{\frac{2}{p}-1} \int_0^1 x^{\frac{1}{p}-1}(1-x)^{\frac{1}{p}-1} dx
= \frac{\Gamma\left(\frac{1}{p}\right)^2}{p^2\Gamma\left(\frac{2}{p}\right)} N^{\frac{2}{p}-1}
$$
Replace $N$ by the list of $p^{th}$ power and sum over it, one will expect
the total number of primitive solutions $\mathcal{N}(p)$ for $x^p + y^p = z^p$ is of the order
$$\mathcal{N}(p) \sim \mathcal{N}_{est}(p) := \frac{\Gamma\left(\frac{1}{p}\right)^2}{p^2\Gamma\left(\frac{2}{p}\right)} \sum_{n=2}^\infty n^{2-p}
= \frac{\Gamma\left(\frac{1}{p}\right)^2(\zeta(p-2)-1)}{p^2\Gamma\left(\frac{2}{p}\right)} 
\tag{*1}$$
Since the sum $\sum\limits_{n=2}^\infty n^{2-p}$ converges for $p > 3$,
we expect $\mathcal{N}(p)$ to be finite. When we plug in some actual numbers, we find
$$\begin{align}
\mathcal{N}_{est}(4) &\sim 0.29893898046807\\
\mathcal{N}_{est}(5) &\sim 0.076793757848265\\
\mathcal{N}_{est}(6) &\sim 0.026448003085251\\
&\;\vdots
\end{align}
$$
They are so small and we don't expect $\mathcal{N}(p)$ to be non-zero at all.
Of course, this argument only works for $p > 3$. I have no idea how to argue for the case $p = 3$.