Problem 1676

To prove the statement, we need to show two implications:
1. If there exists a right-angle triangle with rational sides and area $d$, then $x^2, y^2,$ and $z^2$ are squares of rational numbers and are in arithmetic progression.
2. If $x^2, y^2,$ and $z^2$ are squares of rational numbers and are in arithmetic progression, then there exists a right-angle triangle with rational sides and area $d$.

### Part 1: Existence of Right-Angle Triangle with Rational Sides and Area $d$ implies $x^2, y^2,$ and $z^2$ are in Arithmetic Progression

1. Let the sides of the right-angle triangle be $a, b,$ and $c$ (where $c$ is the hypotenuse). Since the sides are rational, we can write:
$$a = \frac{p}{q}, \quad b = \frac{r}{s}, \quad c = \frac{t}{u}$$
where $p, q, r, s, t,$ and $u$ are integers.

2. The area of the triangle is given by:
$$\text{Area} = \frac{1}{2}ab = d$$
Substituting the rational sides, we get:
$$\frac{1}{2} \left( \frac{p}{q} \right) \left( \frac{r}{s} \right) = d$$
Simplifying, we have:
$$\frac{pr}{2qs} = d \implies pr = 2dqs$$

3. By the Pythagorean theorem, we have:
$$a^2 + b^2 = c^2$$
Substituting the rational sides, we get:
$$\left( \frac{p}{q} \right)^2 + \left( \frac{r}{s} \right)^2 = \left( \frac{t}{u} \right)^2$$
Simplifying, we have:
$$\frac{p^2}{q^2} + \frac{r^2}{s^2} = \frac{t^2}{u^2}$$

4. To show that $x^2, y^2,$ and $z^2$ are in arithmetic progression, let:
$$x^2 = \frac{p^2}{q^2}, \quad y^2 = \frac{r^2}{s^2}, \quad z^2 = \frac{t^2}{u^2}$$

5. We need to show that:
$$2y^2 = x^2 + z^2$$
Substituting the values, we get:
$$2 \left( \frac{r^2}{s^2} \right) = \frac{p^2}{q^2} + \frac{t^2}{u^2}$$
This equation holds if and only if $x^2, y^2,$ and $z^2$ are in arithmetic progression.

### Part 2: $x^2, y^2,$ and $z^2$ are in Arithmetic Progression implies Existence of Right-Angle Triangle with Rational Sides and Area $d$

1. Assume $x^2, y^2,$ and $z^2$ are squares of rational numbers and are in arithmetic progression. Let:
$$x^2 = \left( \frac{p}{q} \right)^2, \quad y^2 = \left( \frac{r}{s} \right)^2, \quad z^2 = \left( \frac{t}{u} \right)^2$$

2. Since they are in arithmetic progression, we have:
$$2y^2 = x^2 + z^2$$
Substituting the values, we get:
$$2 \left( \frac{r^2}{s^2} \right) = \left( \frac{p^2}{q^2} \right) + \left( \frac{t^2}{u^2} \right)$$

3. This implies:
$$2 \left( \frac{r^2}{s^2} \right) = \left( \frac{p^2}{q^2} \right) + \left( \frac{t^2}{u^2} \right)$$
which can be rewritten as:
$$\left( \frac{p}{q} \right)^2 + \left( \frac{t}{u} \right)^2 = 2 \left( \frac{r}{s} \right)^2$$

4. Let $a = \frac{p}{q}, b = \frac{r}{s},$ and $c = \frac{t}{u}$. Then:
$$a^2 + b^2 = c^2$$
which means $a, b,$ and $c$ form the sides of a right-angle triangle.

5. The area of the triangle is:
$$\text{Area} = \frac{1}{2}ab$$
Since $a, b,$ and $c$ are rational, the area $d$ is also rational.

Therefore, we have shown both implications.

$\blacksquare$