[proofplan] We start from a positive integer solution of Fermat's equation at exponent $n \geq 3$. If an odd prime $p$ divides $n$, then writing $n$ as $p$ times a positive integer turns the given solution into a solution at exponent $p$ by raising each base to that positive integer. If no odd prime divides $n$, then elementary divisibility forces $n$ to be divisible by $4$, and the same power-substitution produces a solution at exponent $4$. [/proofplan] [step:Reduce to an odd prime exponent when one divides $n$] Assume first that there exists an odd prime number $p$ such that $p \mid n$. Let $m \in \mathbb{N}$ be defined by $n = pm$. Define positive integers \begin{align*} A := a^m,\qquad B := b^m,\qquad C := c^m. \end{align*} Then \begin{align*} A^p + B^p = (a^m)^p + (b^m)^p = a^{mp} + b^{mp} = a^n + b^n = c^n = c^{mp} = (c^m)^p = C^p. \end{align*} Thus $A,B,C \in \mathbb{N}$ give a positive integer solution of Fermat's equation at the odd prime exponent $p$. [/step] [step:Show that absence of odd prime divisors forces divisibility by $4$] Assume now that no odd prime number divides $n$. We prove that $4 \mid n$. If $n$ were odd, then the set \begin{align*} D := \{d \in \mathbb{N} : d > 1 \text{ and } d \mid n\} \end{align*} would be nonempty because $n \in D$. Let $d_0$ be its least element. If $d_0$ were composite, then $d_0 = rs$ for some integers $r,s$ with $1 < r < d_0$ and $1 < s < d_0$. Since $d_0 \mid n$, this would imply $r \mid n$, contradicting the minimality of $d_0$. Hence $d_0$ is prime. Since $d_0 \mid n$ and $n$ is odd, $d_0$ is an odd prime divisor of $n$, contradicting the present assumption. Therefore $n$ is even. Write $n = 2q$ with $q \in \mathbb{N}$. If $q = 1$, then $n = 2$, contradicting $n \geq 3$. Hence $q > 1$. If $q$ were odd, the same least-divisor argument applied to $q$ would produce an odd prime divisor of $q$, and therefore an odd prime divisor of $n = 2q$, again a contradiction. Thus $q$ is even, so $q = 2m$ for some $m \in \mathbb{N}$. Consequently $n = 4m$, and therefore $4 \mid n$. [/step] [step:Reduce to exponent $4$ when no odd prime divides $n$] By the previous step, there exists $m \in \mathbb{N}$ such that $n = 4m$. Define positive integers \begin{align*} A := a^m,\qquad B := b^m,\qquad C := c^m. \end{align*} Then \begin{align*} A^4 + B^4 = (a^m)^4 + (b^m)^4 = a^{4m} + b^{4m} = a^n + b^n = c^n = c^{4m} = (c^m)^4 = C^4. \end{align*} Thus $A,B,C \in \mathbb{N}$ give a positive integer solution of Fermat's equation at exponent $4$. [/step] [step:Combine the two alternatives] Exactly one of the following alternatives holds: either $n$ has an odd prime divisor, or it has no odd prime divisor. In the first case, the first step gives a positive integer solution at an odd prime exponent. In the second case, the preceding step gives a positive integer solution at exponent $4$. This proves the stated disjunction. [/step]