Originally posted by petrovitchWe are looking for integer solutions for 3.x^2 = 6.y^3
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?
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
x=4
His age is 3 * 4^2 = 48 = 6 * 2^3
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?
Originally posted by wolfgang59X^2 = 2 * Y^3
we are assuming the age is measured in years.
What answers are there if the age is measured in months?
Should Y have any factor F other than 2, then F^6 divides Y^3. For the powers of 2; Y^3 contains an odd number of factors 2 (3, 9, 15, ...).
Y^3 = 2^(6n+3) * F^6
So Y = 2^(2n+1) * F^2
And X = 2^(3n+2) * F^3 with n >= 0 and F = 1 or F > 2.
For F = 1 and n = 0 we get X=4 and Y=2 for an age of 48, already given by previous posters.