Find all positive real numbers x such that both + 1/ and + 1/ are integers.

Let a = + 1/ and b = + 1/.

We suppose that a and b are positive integers.

Then a^{2} = x + 1/x + 2, and

b^{3} = x + 1/x + 3( + 1/) = x + 1/x + 3b.

Hence x + 1/x = a^{2} − 2 = b^{3} − 3b, and so

a^{2} = b^{3} − 3b + 2 = (b + 2)(b − 1)^{2}.

Since a and b are positive integers, this equation holds if, and only if, b + 2 is a perfect square.

That is, if, and only if, b = t^{2} − 2, for some integer t > 1.

Setting y = , we have y + 1/y = t^{2} − 2.

Multiplying by y, we obtain the quadratic equation y^{2} − (t^{2} − 2)y + 1 = 0.

This has solutions y = −1 + ½t(t ± ).

By Viète's formulas, the product of the two roots is 1, and so taking + yields a value of y (and therefore of x) greater than 1, while taking − yields its reciprocal, as we would expect. Note that taking t = 2 yields x = y = 1.

Therefore, all positive real numbers x such that both + 1/ and + 1/ are integers, are given by

x = [−1 + ½t(t ± )]^{3}, where t = 2, 3, 4, ... .

Source: Inspired by a puzzle posted by Mark Nandor