Introduction to Fermat Numbers and Their Unique Properties

Fermat numbers are a fascinating sequence in number theory, defined by the formula (F_n 2^{2^n} 1) where (n) is a non-negative integer. These numbers are notable for their unique properties and the challenges they present in terms of prime factorization. In this article, we will explore how to determine the prime factors of coprime squares, and why the left and right sides of a specific equation have different properties that lead to a proof of Fermat's theorem.

Understanding Fermat Numbers and Their Characteristics

The Fermat numbers, (F_n 2^{2^n} 1), are a sequence of numbers that have intrigued mathematicians due to their properties and the challenges associated with factorizing them. One of the key characteristics of Fermat numbers is that their prime factors are of a special form. In particular, the prime factorization of a sum of coprime squares can only include primes of the form (1 mod 4).

Proof That the Right Side Admits Only 1 Mod 4 Primes

Consider the equation (2^a - 2^b 2^{2^c}) where (a geq b geq 1). First, observe that if (c 0), the left side equals (12) while the right side is (2^0 1), which is a contradiction. If (c 1), we get (2^b - 2^2 2), leading to (2^{b-2} - 1 1), which simplifies to (2^{b-2} 2), i.e., (b 3). For (c geq 2), we note that the left side is of the form (2^a - 1), which implies (2^a - 1 equiv 3 mod 4).

Prime Divisors of Fermat Numbers

Since (2^a - 1 equiv 3 mod 4), there must be a prime divisor (q) of (2^a - 1) of the form (4k 3). This means that (q) must also be a divisor of (2^{2^c} - 1), which is a sum of two squares. However, if (q mid m^2n^2) for some integers (m) and (n), then (q mid m) and (q mid n). This is because (q) is a prime of the form (4k 3) and it cannot divide the product of any squares except for (m) and (n) themselves. This leads to a contradiction, as (q) cannot divide any other part of the equation.

No Solutions for (c geq 1)

From the above analysis, we conclude that there can be no solutions for (c geq 1), thus confirming that the equation (2^a - 2^b 2^{2^c}) does not hold for (c geq 1).

Proving the Uniqueness of Solutions for Special Cases

For the case when (c 0), the only solution is ({ab} {1, 2}), which gives us (120210).

Formulating the General Equation and Simplifying

We consider the equation (2^{ab} - 2^a - 2^b 2^{2^c}). Moving terms to the right side, we obtain (2^{ab} 2^a2^b 2^{2^c}). Let (t ab), (x a), (y b), and (z 2^c). Then, the equation becomes (2^t 2^x(2^{y-x} 2^z)).

Analyzing the Equation with Symmetry

Assuming (x leq y leq z), dividing by (2^x) yields (2^{y-x}2^{z-x} 2^{t-x}). The only way the parities of the left and right sides match is if the three terms are odd-odd-odd or odd-even-odd. The second case is the only one that works, so we have (12^{z-x} 2^{t-x}). This implies (y x) and (z 1 t - x2).

Setting permutations of (ab2^c) equal to (xxx1x2) reveals only two solutions. This concludes the proof that the only solutions are ({120210}).