I promise to explain my "secret" algorithm (but there is no great secret...) when it will be proved enough good and optimized to successfully answer to the 5x5 question.

Christian

Hello folks, I don't know how Christian does it, but to take 3x3 as an example, I'd start with a generalization of all 3x3 multiplicative magic squares, such as: z*x z*x^2*y^2 z*y z*y^2 z*x*y z^x^2 z*x^2*y z z*x*y^2 The magic product is then x^3*y^3*z^3 = 216 when z = 1, x = 2, y = 3, the smallest integer assignmments. Stepwise incrementing of these values while skipping those results showing repeated entries will then yield squares with the remaining products on Christian's list. Ditto for 4x4. But I must say, 5x5 looks tricky. Lee