How to explain to a layperson why Fermat's Last Theorem involves non-trivial math? Fermat's Last Theorem states, given$$x^n + y^n = z^n$$ no three integers $x,y,z$ will satisfy the equation given integer value of $n$ greater than two.
On the surface this seems like something that can be posed to a middle school student, whereby they will try a few cases for different values of $n$ and conclude that as $n$ goes large, there probably wouldn't exist a triple $x,y,z$ such that the equation will be satisfied.
What would be the first step of transforming this question to involve a non-trivial application of ring theory or group theory that Andrew Wiles' proof demonstrated? In other words, how do mathematicians see the connection between this equation and higher branches of mathematics?
Are there similar examples i.e. estimating the next prime?
 A: I will try to give you a VERY quick overview on the strategy of the proof of FLT. Of course I cannot avoid to use very technical tools, such Galois representation. If you don't know about these stuff, I hope you can at least follow the "shape" of the argument.
The starting observation is the following: for $n\in\mathbb N$, let FLT($n$) be the statement "there are no triples $(a,b,c)$ of integers with $abc\neq 0$ such that $a^n+b^n=c^n$". Then it is elementary to see that FLT($d$) implies FLT($n$) whenever $d\mid n$. Therefore it is sufficient to prove FLT($p$) for every odd prime $p$ and for $n=4$ in order to prove FLT($n$) for all positive integers $n\geq 3$. The cases $p=3$ and $n=4$ were already known to Euler, I believe, and can be proven by elementary methods, so we can assume that $p\geq 5$.
Now the non-elementary math comes in. Suppose that there is a triple of integers $(a,b,c)$ which contradicts FLT($p$) for some prime $p$. One can construct the following elliptic curve over $\mathbb Q$:
$$E_{a,b,c}\colon y^2=x(x-a^p)(x-b^p)$$
It is possible to show (assuming wlog that $(a,b,c)$ is a coprime triple, $a\equiv -1\bmod 4$ and $2\mid b$) that this is a semistable elliptic curve whose conductor is $\prod_{l\mid abc}l$.
Moreover, Serre and Frey proved the following theorem: let
$$\overline{\rho}_{a,b,c}\colon \text{Gal}(\overline{\mathbb Q}/\mathbb Q)\to GL_2(\mathbb F_p)$$
be the residual Galois representation at $p$ attached to $E_{a,b,c}$. Then:


*

*$\overline{\rho}_{a,b,c}$ is absolutely irreducible;

*$\overline{\rho}_{a,b,c}$ is odd;

*$\overline{\rho}_{a,b,c}$ is unramified outside $2p$ and flat at $p$


The key idea is then the following: suppose that we can show that for every (semistable) elliptic curve $E$ over $\mathbb Q$ the representation $\overline{\rho}_E$ is the residual representation of the $p$-adic Galois representation attached to a weight $2$ newform $f_E$. Then we can apply a theorem of Ribet, using the properties of $\overline{\rho}_{a,b,c}$, to show  that for $\overline{\rho}_{a,b,c}$ we can choose such a newform in $S_2(\Gamma_0(2))$. But then we have a big problem: this space is $0$-dimensional! Therefore we are led to a contradiction, and there cannot be any triple $(a,b,c)$ contradicting FLT($p$).
This whole argument was already known long before Wiles' proof. But the thing that was missing was the proof of the (nowaday) so-called modularity theorem: every elliptic curve over $\mathbb Q$ is modular, i.e. its $p$-adic Galois representation is the $p$-adic Galois representation attached to a weight $2$ newform.
Even though this was the only missing part, it is by far the most difficult one! Wiles was able to prove it for semistable curves, which was enough for proving FLT, but some years later the result has been improved to all elliptic curves over $\mathbb Q$ by Breuil, Conrad, Diamon and Taylor.
Anyway, if you are interested in more details, you can read the (amazing) first chapter of "Modular forms and Fermat's last theorem", by Cornell, Silverman and Stevens eds. Of course it is a math textbook, so you need some background in these type of topics in order to understand it.
A: Actually this is a good opportunity to explain to the layman how mathematics works, but I would change the motivating example a little bit.
Since we are searching for integer or rational solutions to the equation $x^n+y^n=z^n$, I would take a simple Diophantine equation and explain how to solve it using simple methods, for example parity (property of sums of integers), then change a little bit the equation in a way that our method fails, now we need a fancier method, then I would take yet another equation such that our last method fails too. At this point I would make a parallel explanation regarding why $x^n+y^n=z^n$ is a hard problem to solve.
PS$_1$: It's not hard to find problems in Diophantine equations such as the ones I talk above.
PS${_2}$: If your audience has a little bit of experience with mathematics you could actually explain why $x^n+y^n=z^n$ is hard using factorization techniques.
