# Fermat's Magic Cube

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In 1640, Pierre de Fermat sent a letter to Marin Mersenne about an order-4 magic cube with 64 magic sums out of 76 possible. In a perfect magic cube the rows, columns, pillars, space diagonals, and the diagonals of each *n*×*n* orthogonal slice sum to the same number. For the order-4 case, numbers 1 to 64 are required. In 1972, Richard Schroeppel proved that a perfect order-4 magic cube was impossible (see Details).

Contributed by: Ed Pegg Jr (March 2011)

Open content licensed under CC BY-NC-SA

## Snapshots

## Details

Lemma 1: In a magic square of order 4, the sum of the corners is .

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 ; ; .

Let (= 130) be the sum of a row. Consider a corner . There are three corners connected by an edge to . Each must have the same value , contradicting the requirement that all values are different. Thus, there is no magic cube of order 4.

QED. (Proof by Richard Schroeppel, 1972)

## Permanent Citation