Said a chess master playing for free, "My age I've divided by three. The result is a square. Yet my age, I declare, is six times a cube. How can that be?

Originally posted by petrovitch Said a chess master playing for free, "My age I've divided by three. The result is a square. Yet my age, I declare, is six times a cube. How can that be?

We are looking for integer solutions for 3.x^2 = 6.y^3

ie x^2 = 2.y^3

A set of solutions is

x = 2^(3n-1) and y=2^(2n-1)

The only solution giving a reasonabe age for our master is n=1

48 is the only reasonable age. 2x2x2x6, which fits the criteria for 1/3 being a square (4x4=16).
The next number is 3x3x3x6=168. 1/3rd of that is 56, which is not a square.
I doubt there are any chess masters much over 80, let alone older than 168. If there were, would they play for free?