To prove FLT, it suffices to prove it for any prime $n \ge 5$.

I once read somewhere, (can't find link) that to prove Fermat's Last Theorem, assuming it has been proven for $n = 3, 4$, it suffices to prove it for every prime $n \ge 5$. I have no idea why this is true. Can somebody explain?

• What is FLT???? – JPi Mar 20 '14 at 22:06
• $a^{pq}+b^{pq}=(a^q)^p+(b^q)^p$ – Ian Mateus Mar 20 '14 at 22:07
• @JPi: Fermat's Last Theorem. – Asaf Karagila Mar 20 '14 at 22:07

Suppose $n$ is not prime, then it can be written as $pq$ for some $p,q\in\mathbb{N}$, with $q$ prime, and $a^n$ can be written as $(a^p)^q$ so if there is a solution for $n$ there also is one for $q$ (which is smaller than $n$).
• The OP is asking why it suffices to prove it for primes greater than $4$. – user71641 Mar 20 '14 at 22:06
• $a^{p^q} \neq \left(a^p\right)^q$. – Eric Towers Mar 20 '14 at 22:30