# Solution to puzzle 138: Integer sum of roots

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 a2 = x + 1/x + 2, and
b3 = x + 1/x + 3( + 1/) = x + 1/x + 3b.
Hence x + 1/x = a2 − 2 = b3 − 3b, and so
a2 = b3 − 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 = t2 − 2, for some integer t > 1.

Setting y = , we have y + 1/y = t2 − 2.
Multiplying by y, we obtain the quadratic equation y2 − (t2 − 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