This week’s Classic from Dean Ballard asks about placing prime numbers on the corners of a cube such that the 4 corners on every face add up to the same number. First we will look at general solutions and then take a look at solutions for primes:

We label the numbers on the corners of a cube somewhat arbitrarily with letters [a,b,c,d,e,f,g,h]:

We notice that if the face sums are equal we get a set of six simultaneous equations relating pairs of corners of faces with a shared edge:

a+b=f+g; c+d=e+h; c+f=a+h; d+g=b+e, b+c=g+h, a+d=e+f

Solving these simultaneously we find there is no unique solution but a family of solutions with the requirement that all the long diagonal corner differences be equal:

Solution Condition: (a-f) = (e-d) = (g-b) = (c-h ) .

We can find a non-prime solution quickly with the arithmetic progression [0, 1, 2, 3, 4, 5, 6, 7] assigned as [a,b,c,d,e,f,g,h] = [3, 4, 7, 0, 1, 2, 5, 6]. Permuting corner pairs also produces valid solutions. Each long diagonal corner difference is 1 and the face sums are 18. This is interesting because it implies that any arithmetic progression can be a solution to the general non-prime puzzle since we can obviously multiply all the corners of a valid solution or add a constant and still have a valid solution. However there are many non-arithmetic progressions that form valid solutions as well.

Looking for primes we can quickly see a solution with [a,b,c,d,e,f,g,h] =[13, 17, 31, 3, 5, 11, 19, 29] or when sorted [3, 5, 11, 13, 17, 19, 29, 31] . The long diagonal corner differences are 2 and the face sums are 64. There is a solution nearby with [a,b,c,d,e,f,g,h] = [13, 17, 31, 5, 7, 11, 19, 29] or when sorted [5, 7, 11, 13, 17, 19, 29, 31]. The long diagonal corner differences are 2 and the face sums are 66.

For further prime solutions we can hunt for them (the twin prime conjecture implies there are an infinite number of prime pairs with any even gap) or even use our observation above that all arithmetic progressions are solutions. According to Green-Tao there will also be a very large (infinite?) number of solutions of this type. Looking in Wikipedia we see the following table for prime arithmetic progressions:

Choosing values for k=8 terms we find a solution [a,b,c,d,e,f,g,h] = [829, 1039, 1669, 199, 409, 619, 1249, 1459] or when sorted [199, 409, 619, 829, 1039, 1249, 1459, 1669]. The long diagonal corner differences are 210 and the face sums are 3736.