Frank Rubin asks this interesting problem: --------------------------------- "It is possible to write a square on each of the 6 faces of a cube (such as a die) so that the 3 faces surrounding each of the 8 vertices sum to a square. For example, write 1 on two opposite faces, and 4 on the other 4 faces. The sum of the 3 faces surrounding each vertex is then 1+4+4=9, a square. Can this be done using the squares of 6 distinct whole numbers? If so, what set of 6 such squares has the smallest sum?" --------------------------------- As for the "3x3 magic squares of squares" unsolved problem, it is a system of 8 equations summing 3 squares. If the die is composed by the 6 integers: a² b², c², d², e² f² (couples a²/f², b²/d² and c²/e² being on opposite faces) the 8 equations are: a²+b²+c² = s² a²+c²+d² = t² a²+d²+e² = u² a²+e²+b² = v² f²+b²+c² = w² f²+c²+d² = x² f²+d²+e² = y² f²+e²+b² = z² I am able to find numerous solutions of 7 equations on 8, for example: 0²+132²+176² = 220² 0²+176²+468² = 500² 0²+468²+351² = 585² 0²+351²+132² = 375² 1200²+132²+176² = 1220² 1200²+176²+468² = 1300² 1200²+468²+351² = 1335² 1200²+351²+132² = 1580625 = 75² x 281 = (1257.229...)² But I am unable to find a complete solution of the 8 equations. Do you see any mathematical impossibility? Christian.