Think Outside the Box. A 4-dimensional Box

The solution to this problem hinges on this lovely theorem due to Jacobi.

First, what does the Jacobi four-square theorem state? It says that, for a positive odd integer $n$ , the number of representations of $n$ as a sum of 4 squares of integers is given by $8\sigma_{0}(n)$ . This counts permutations as distinct, eg $1^2+0^2+0^2+0^2$ and $0^2+1^2+0^2+0^2$ are counted as two different representations. (There is another explicit closed form for even $n$ but that isn't necessary here.)

Second, we need to find an appropriate $n$ . Since the hypersphere has radius $r = 7$ and we know that a 4D hypersphere is described by the equation $x^2+y^2+z^2+w^2 = r^2$ , we have our $n = r^2 = 49$ . This is especially convenient since 49 has only the divisors 1, 7, and 49.

Therefore, $8\sigma_{0}(49) = 8(1+7+49) = \boxed{456}$ .

Just for the information, the given shape is also called a $glome$ .

More elementary, we write $7^2$ as the sum of four perfect squares:

• $49 = 0^2 + 0^2 + 0^2 + 7^2$ . The 7 may be along each the four axes, and may be positive or negative. This gives 8 solutions.

• $49 = 0^2 + 2^2 + 3^2 + 6^2$ . The non-zero values may be positive of negative, giving 8 possible choices. They may be permuted in $4! = 24$ ways. Therefore we have another 8 x 24 = 192 solutions.

• $49 = 2^2 + 2^2 + 4^2 + 5^2$ . This gives 16 x 12 = 192 solutions.

• Finally, $49 = 1^2 + 4^2 + 4^2 + 4^2$ . This results in 16 x 4 = 64 solutions.

You can conclude that there are no other possibilities simply be checking the solutions. (Hint: At least one of the terms must be greater than $3^2$ . Systematically work from the highest to the lowest value.)

Adding everything together, we get $\boxed{456}$ solutions.