The problem asks you to show that the point
$\left( \frac{m^2 - n^2}{m^2 + n^2}, \frac{2mn}{m^2 + n^2} \right)$
lies on the unit circle, given that $m$ and $n$ are integers and not both zero. Let's solve it step-by-step.

Proof:

A point $(x, y)$ lies on the unit circle if it satisfies the equation: $x^2 + y^2 = 1.$

We are given: $x = \frac{m^2 - n^2}{m^2 + n^2}, \quad y = \frac{2mn}{m^2 + n^2}.$

Now, let's compute $x^2 + y^2$ to verify if it equals 1.

1. Compute $x^2$: $x^2 = \left( \frac{m^2 - n^2}{m^2 + n^2} \right)^2 = \frac{(m^2 - n^2)^2}{(m^2 + n^2)^2}.$ Expanding $(m^2 - n^2)^2$: $(m^2 - n^2)^2 = m^4 - 2m^2 n^2 + n^4.$ So: $x^2 = \frac{m^4 - 2m^2 n^2 + n^4}{(m^2 + n^2)^2}.$

2. Compute $y^2$: $y^2 = \left( \frac{2mn}{m^2 + n^2} \right)^2 = \frac{4m^2 n^2}{(m^2 + n^2)^2}.$

3. Sum $x^2 + y^2$: $x^2 + y^2 = \frac{m^4 - 2m^2 n^2 + n^4}{(m^2 + n^2)^2} + \frac{4m^2 n^2}{(m^2 + n^2)^2}.$ Since the denominators are the same, we can combine the numerators: $x^2 + y^2 = \frac{m^4 - 2m^2 n^2 + n^4 + 4m^2 n^2}{(m^2 + n^2)^2}.$ Simplify the numerator: $m^4 - 2m^2 n^2 + n^4 + 4m^2 n^2 = m^4 + 2m^2 n^2 + n^4 = (m^2 + n^2)^2.$ So: $x^2 + y^2 = \frac{(m^2 + n^2)^2}{(m^2 + n^2)^2} = 1.$

Conclusion:

Since $x^2 + y^2 = 1$, the point
$\left( \frac{m^2 - n^2}{m^2 + n^2}, \frac{2mn}{m^2 + n^2} \right)$
lies on the unit circle.

Do you have any questions or would you like further clarification?

Related Questions:

1. What is the geometric meaning of the unit circle?
2. How does this parametrization relate to trigonometric identities?
3. Can this point be interpreted as a rational parameterization of the unit circle?
4. What happens if both $m$ and $n$ are zero?
5. How is this approach connected to Pythagorean triples?

Tip:

This parametrization is related to rational points on the unit circle, often linked with Pythagorean triples when $m$ and $n$ are integers.