Fermat's Magic Cube
![]() Proof: Add together each edge of the square and the two diagonals. This covers the square entirely, and each corner twice again. This adds to , so twice the corner sum is .Lemma 2: In a magic cube of order 4, the sum of any two corners connected by an edge of the cube is .Proof: Call the corners and . Let , , and , be the corners of any two edges of the cube parallel to . Then , , and are all the corners of magic squares. So ; ; .![]() "Fermat's Magic Cube" from The Wolfram Demonstrations Project http://demonstrations.wolfram.com/FermatsMagicCube/ Contributed by: Ed Pegg Jr | ||||||||||||||
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