The Diophantine quadratic equation you provided is:

$x^2 + y^2 = z^2$

This is known as Pythagoras' equation, and its solutions are called Pythagorean triples. The general integer solutions to this equation can be written in parametric form. These solutions are:

$x = m^2 - n^2$ $y = 2mn$ $z = m^2 + n^2$

where $m$ and $n$ are integers, and $m > n > 0$. These triples $(x, y, z)$ are called primitive Pythagorean triples if $m$ and $n$ are coprime (i.e., their greatest common divisor is 1) and one of them is even.

Explanation:

• The expressions for $x$, $y$, and $z$ generate all possible primitive solutions to the equation $x^2 + y^2 = z^2$.
• Non-primitive solutions (i.e., multiples of the primitive ones) can be generated by multiplying the primitive triples by a common factor $k$. So the general solution for all Pythagorean triples is given by:

$x = k(m^2 - n^2), \quad y = k(2mn), \quad z = k(m^2 + n^2)$ where $k$ is any positive integer.

Example:

For $m = 3$ and $n = 2$, we get the primitive Pythagorean triple:

$x = 3^2 - 2^2 = 9 - 4 = 5$ $y = 2 \cdot 3 \cdot 2 = 12$ $z = 3^2 + 2^2 = 9 + 4 = 13$

Thus, $(5, 12, 13)$ is a Pythagorean triple.

Would you like further details or have any questions about the solution?

Related Questions:

1. How do you prove that these parametric solutions satisfy $x^2 + y^2 = z^2$?
2. Can there be negative values for $x$, $y$, or $z$ in the solutions?
3. What is the significance of coprimeness between $m$ and $n$ in generating primitive triples?
4. How do you find the greatest common divisor (GCD) between two numbers?
5. How can we find the number of primitive triples for a given range of $z$?

Tip:

To verify if a set of numbers $(x, y, z)$ forms a Pythagorean triple, just check whether $x^2 + y^2 = z^2$ holds true.