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Solution to puzzle 138: Integer sum of roots

Skip restatement of puzzle.Find all positive real numbers x such that both root x + 1/root x and cube root of x + 1/cube root of x are integers.


Let a = root x + 1/root x and b = cube root of x + 1/cube root of x.
We suppose that a and b are positive integers.

Then a2 = x + 1/x + 2, and
b3 = x + 1/x + 3(cube root of x + 1/cube root of x) = 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 = cube root of x, 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 ± square root of (t^2 - 4)).
By Viète's formulas, the product of the two roots is 1, and so taking +square root of (t^2 - 4) yields a value of y (and therefore of x) greater than 1, while taking square root of (t^2 - 4) 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 root x + 1/root x and cube root of x + 1/cube root of x are integers, are given by
x = [−1 + ½t(t ± square root of (t^2 - 4))]3, where t = 2, 3, 4, ... .

Source: Inspired by a puzzle posted by Mark Nandor

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