Fermat’s Last Theorem concerns the case where n > 2. However, we can start from n=2 to derive the general expression for Pythagorean triples and obtain the original ideas for proving Fermat’s Last Theorem.
Coprime Solutions
We seek solutions within the rational integers, denoted by \mathbb{Z}. To solve the Diophantine equation x^2 + y^2 = z^2, we first note that it is a homogeneous equation. This implies that if a solution exists, we can divide by the greatest common divisor to obtain a set of pairwise coprime solutions. In other words, if a solution exists, a coprime solution must exist. This is a general property of solutions to x^n + y^n = z^n. Let us assume that (x, y, z) = (a, b, c) is a coprime solution to the equation x^2 + y^2 = z^2.
Next, we analyze the congruence properties. We choose the prime factor 2, which means analyzing the parity of a, b, c. In \mathbb{Z}, 2 is a special prime because it is the prime with the smallest absolute value, resulting in the fewest congruence cases (any integer divided by 2 either leaves a remainder of 1 or is divisible). It is easy to notice the following fact:
If n \equiv 1 \pmod 2, then n^2 \equiv 1 \pmod 4.
From this, we can deduce that one of a and b must be even and the other must be odd. The proof is as follows: since it is a coprime solution, they cannot both be even. If they were both odd, then a^2 \equiv 1 \pmod 4 and b^2 \equiv 1 \pmod 4, which gives (a^2 + b^2) \equiv 2 \pmod 4. However, a^2 + b^2 = c^2, so c^2 \equiv 2 \pmod 4. This is a contradiction, as a square number modulo 4 can only be 0 or 1.
Since one of a, b is even and the other is odd, c must be odd. We can set a to be even and b to be odd. Note that: a^2 = c^2 - b^2 = (c - b)(c + b) We know that the common divisor of m and n must also divide m + n and m - n. Therefore, the common divisor of c - b and c + b must also divide 2c and 2b. Since c and b are both odd, c - b and c + b have a common divisor of 2. Thus, we can set c - b = 2u, c + b = 2v, and a = 2\alpha, which gives: \alpha^2 = uv Here u + v = c and v - u = b. Since b and c are assumed to be coprime, it follows that (u, v) = 1. If the product of two coprime numbers is a square, then these two numbers are themselves squares, up to a unit factor. It should be noted that the phrase “up to a unit factor” is indispensable. The units in \mathbb{Z} are \pm 1, which tells us that 6 can be written not only as 2 \times 3, but also as (-2) \times (-3). Therefore, we can set u = \pm p^2 and v = \pm q^2, yielding the general formula for primitive Pythagorean triples: a = \pm 2pq, \quad b = \pm q^2 \mp p^2, \quad c = \pm q^2 \pm p^2
Review of Steps
Let us briefly analyze each of our steps and consider which ones can similarly be used to analyze solutions when n > 2. First, the step stating that if a solution exists, a pairwise coprime solution must exist, holds for any n. Second, we used congruence analysis regarding the prime 2; this step does not apply directly when n > 2. However, after extending the number field and choosing other prime factors, this step remains feasible. Finally, we used a crucial property: “If the product of two coprime numbers is a square, then these two numbers are themselves squares, up to a unit factor.” If we replace the square with a cube, a fourth power, etc., the conclusion still holds, which is related to the unique factorization property of \mathbb{Z} (the Fundamental Theorem of Arithmetic).
We also see that to solve x^2 + y^2 = z^2, we factor it as x^2 = (z - y)(z + y), and for the sake of analytical convenience, it is factored into a product of linear polynomials. However, for n > 2, this cannot be achieved within the realm of real numbers. To achieve this, it is necessary to extend the integers to the realm of complex numbers, which demonstrates the necessity of extending the number field. However, in the extended “integers,” many properties that are relatively obvious in the rational integers no longer hold, such as unique factorization. This requires further techniques to overcome these difficulties, which will be discussed in subsequent articles.
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