# Is there a simple way to illustrate that Fermat's Last Theorem is plausible?

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?

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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 $p = 3$.

Is there a simple way to demonstrate that Fermat’s Last Theorem is (at least) plausible? Probably not, other than computer searches that keep coming up negative. What Fermat had, or thought he had, is, as far as I know, a total mystery. The famous note in the margin was found years after his death and there does not seem to be other work of his on the matter.

has trying to proof FLT always been ‘a shot in the dark’ for everyone anyway? Not at all. Quite a lot of modern algebra (e.g., ideals) arose out of very structured (even if unsuccessful) attempts to solve FLT. Proofs for particular exponents were established, giving at least partial proofs, and there were not just shots at the dark. Further studies of the problem led to connections with elliptic curves, and finally to Wiles’ proof.

Just to compare FLT to another famous problem, the twin-primes conjecture, I believe it is safe to say that FLT produced a lot of important theories that eventually did not lead to a proof of FLT, but are very important. The twin-primes conjecture on the other hand did not give rise to plenty of theories and attempted proofs of it were somewhat more of a shot in the dark. It is in some sense not surprising that FLT was solved, especially once the connection with elliptic curves was made (not that I’m saying there is anything trivial in the actual proof, just that once a connection with something as diverse and important as elliptic curves is made, one expects the rich setting to allow for a proof). The twin-prime conjecture though, that now seems to be on the verge of being solved, would have remained ‘impossible’ if it were not for the sudden leap made by Yitang Zhang.

For each n, you will ultimately arrive at a product of n quantities equaling a product of exactly two quantities. But due to various co-primality & positivity conditions previously imposed, this will only be possible for $n=2$. All you need is Newton’s binomial theorem and/or some basic knowledge of modular arithmetic. Solving the cases $n=2$ and $n=3$, and then contrasting them, is particularly enlightening insofar intuition is concerned.