Problem 487

Solution:

Let
$$y = \frac{x^{2} - 1}{2} \quad \text{and} \quad z = \frac{x^{2} + 1}{2}.$$
Because $x$ is odd, $x^{2} - 1$ and $x^{2} + 1$ are both even and therefore $y$ and $z$ are integers. Moreover, because $x$ is more than $1$, $x^{2} - 1$ and $x^{2} + 1$ are more than $0$ and thus $y$ and $z$ are positive. Finally, the desired equation $x^{2} + y^{2} = z^{2}$ is equivalent to
$$\begin{aligned} x^{2} + \left(\frac{x^{2} - 1}{2}\right)^{2} & = \left(\frac{x^{2} + 1}{2}\right)^{2} \\ x^{2} + \frac{\left(x^{2} - 1\right)^{2}}{4} & = \frac{\left(x^{2} + 1\right)^{2}}{4} \\ 4x^{2} + \left(x^{2} - 1\right)^{2} & = \left(x^{2} + 1\right)^{2} \\ 4x^{2} + x^{4} - 2x^{2} + 1 & = x^{4} + 2x^{2} + 1, \end{aligned}$$
which is true.