Solve the Diophantine equation x² + y² = z² in positive integers (Pythagorean triples).

Step 1: Introduce Primitive Pythagorean Triples

We are looking for positive integer solutions to the equation $x^2 + y^2 = z^2$. These are known as Pythagorean triples. A common approach is to first find primitive Pythagorean triples, where $x$, $y$, and $z$ are coprime. All other Pythagorean triples can be generated by multiplying a primitive triple by a common integer factor. The general form for primitive Pythagorean triples is given by Euclid's formula, where $m$ and $n$ are coprime positive integers of opposite parity, and $m > n$.

Step 2: Verify the Formula

Let's substitute the expressions for $x$, $y$, and $z$ into the original equation to verify that they satisfy the Pythagorean theorem. This step confirms the validity of Euclid's formula.

Step 3: Expand and Simplify

Expanding the squares on the left side of the equation, we get $(m^4 - 2m^2n^2 + n^4)$ from $(m^2 - n^2)^2$ and $4m^2n^2$ from $(2mn)^2$. The right side is $(m^2 + n^2)^2 = m^4 + 2m^2n^2 + n^4$.

Step 4: Combine Terms

Combining the terms on the left side, $-2m^2n^2 + 4m^2n^2$ simplifies to $2m^2n^2$. This makes the left side equal to the right side, confirming the identity.

Step 5: General Solution

The general solution for all Pythagorean triples (not just primitive ones) is obtained by multiplying the primitive triple formulas by an arbitrary positive integer $k$. Here, $m$ and $n$ are coprime positive integers, $m > n$, and $m$ and $n$ have opposite parity. $k$ is any positive integer.

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