One day I just happened to wake up thinking of Fermat’s last theorem and an urge to try and get a feel for why it is plausible at all. I don’t have the math chops to grok Wiles’ proof, but was just reaching for something which could suggest to me why it might be true at all. This is what I came up with.

The statement

\(a^n+b^n=c^n\) has no non-trivial solutions for \(a,b,c \in \mathbb{N}\) for \(n>2\).

A “trivial solution” is where either \(a\) or \(b\) is zero.

Why should this be true?

Without loss of generality, let’s assume \(a > b\). Suppose we look at \(a\), \(b\) and \(c\) in their digit representations modulo some \(m\) (like 10). Denote the number of digits (without leading zeros) of a number \(k\) in base \(m\) as \(|k|_m\). For now, we can omit the \(m\) subscript since we can put it back in later.

We have \(|a^n| \approx n|a|\) and \(|a^n| + \{0,1\} = |c^n|\).

If we take the task of determining the numbers \(a\), \(b\) and \(c\) to be the task of determining their digits, it looks like we have \(\approx 3|a|\) variables and \(\approx n|a|\) equations or constraints on them. So it seems reasonable that no solutions might be possible for \(n > 3\). ¹

This intuition would also suggest that the generalized version of the theorem for \(a^m + b^n = c^k\) would also not have solutions for \(\text{min}(m,n,k) > 3\).

More carefully?

No need :D I’m pretty satisfied with this hand wavy argument for why FLT just might be true. Anyone interested in making a more careful statement here?

\(|a^n|\) lies in the range \(n(|a|-1)+1\) to \(n|a|\) and therefore the number of constraints can also be anywhere in this range. So if we look for \(n(|a|-1)+1 > 3|a|\), or \(n|a| > 3|a|+n-1\) that can be met with \(n > 3\) as well.

The nature of the constraints

The “more constraints than variables” heuristic was a tad too hand wavy for taste. So at first I tried to see if changing what “multiplication” means by defining it to not do a carry over (and similarly addition to not use carry over), would FLT hold then as well. For example, if addition were defined to be digit-wise addition modulo the base and multiplication were defined to be polynomial multiplication where the resultant coefficients are also considered modulo the base (without carry over), this meets the criteria for a ring, which the integers also meet. However this notion of multiplication is not sufficient for the heuristic to work, and we can indeed find many solutions to FLT in this ring for n >= 3. Given such a convolution based multiplication without carry, we have \(|ab| = |a| + |b| - 1\). So \(|a^2| = 2|a|-1\), … \(|a^n| = n(|a|-1)+1\). In this case, information is lost when collecting coefficients in the multiplication process and therefore the constraints are correspondingly less tight. However, I still wonder if there is a \(k\) such that \(\forall n > k\) no solutions exist. This should be checkable with a little more reasoning work and properties of the algebraic structure such as existence of inverses will also be relevant.

However, if we consider the theorem in the context of the ring of polynomials in a single variable - i.e. \(a = \sum_{k\geq0}{a_kx^k}\), \(b = \sum_{k\geq 0}{b_kx^k}\) and \(c = \sum_{k\geq 0}{c_kx^k}\), where \(a_k,b_k,c_k \in \mathbb{Z}\) with addition being ordinary polynomial addition and multiplication being ordinary polynomial multiplication, we can expect FLT to hold since information is not destroyed when collecting coefficients in the multiplication unlike in the “modulo base” case mentioned above.

(27 Sep 2025) This document https://www.columbia.edu/~abb2190/FLTPoly.pdf discusses the case of the polynomial ring (albeit \(\mathbb{C}[x]\) and provides 3 proofs for FLT for that case. That it is true for non-constant polynomials in \(\mathbb{C}[x]\) implies it ought to be true for non-constant polynomials with integer coefficients as well. A personal TODO item is to try to get a deeper feel for “Proof 3: Geometry”.