Worked solution: Wiles's modularity proof via the Taylor–Wiles method (1994)

Fermat's Last Theorem (FLT) states that for every integer $n > 2$, there are no nonzero integers $x, y, z$ satisfying $x^n + y^n = z^n$. As recorded in Wiles's introduction (Wiles 1995, p. 443) and reviewed in Qiu et al. (2025, Section 3.1.1, Theorem 3.1), it suffices to prove the theorem when the exponent is $n = 4$ or an odd prime $p \ge 5$, and when the three integers are pairwise coprime.

Why does this reduction hold? Every integer $n > 2$ is either a power of $2$ (so $4$ divides $n$, say $n = 4m$) or divisible by at least one odd prime $p$ (say $n = pm$). If $x^n + y^n = z^n$, then in the first case $(x^m)^4 + (y^m)^4 = (z^m)^4$ is a solution for exponent $4$, and in the second case $(x^m)^p + (y^m)^p = (z^m)^p$ is a solution for the odd prime $p$. Fermat proved the case $n = 4$ around 1667 by his method of infinite descent, and Euler proved $p = 3$ between 1753 and 1770 (with a gap later filled by Legendre; see Qiu et al. 2025, Section 2.1). Moreover, if any two of $a, b, c$ shared a prime factor $q$, then $q^p$ would divide the third as well, so we may divide out $\gcd(a, b, c)^p$ until $a, b, c$ are pairwise coprime.

This reduction sets the stage for the modern proof: we assume for contradiction that there exists a prime $p \ge 5$ and nonzero pairwise coprime integers $a, b, c$ with $a^p + b^p = c^p$, and we use this hypothetical triple to build a geometric object in the next step.