# A cube of primes

This week’s Riddler Classic is a question about prime numbers.

Consider a cube, which has eight vertices, or corners. Suppose I assign a prime number to each vertex. A “face sum” is the value I get when I add up all four prime numbers on one of the six faces.

Can you find eight distinct primes and arrange them on a cube so that the six face sums are all equal?

Extra credit: Can you find another set of eight distinct primes that can similarly be arranged on the vertices of a cube? How many more can you find?

Extra Extra credit: Same puzzle for the other four platonic solids.

Here is my solution:
[Show Solution]

## 5 thoughts on “A cube of primes”

1. Glnc says:

Look into Dirichlets theorem on arithmetic progressions. It resolves your question about whether there are infinitely many solutions

1. Can you elaborate? Dirichlet’s theorem says that for any $a$, $b$ coprime, there are infinitely many primes of the form $a + nb$. So for example, there are infinitely many primes of the form $4+3n$.

Dirichlet’s theorem tells me that there are arithmetic progressions that contain infinitely many primes, but it does not tell me that there are infinitely many arithmetic progressions of primes, which is what I would need to guarantee infinitely many solutions to the problem.

1. I think the Green-Tao Theorem is closer to what we need here — but thank for pointing me in the right direction! I just assumed we couldn’t prove anything because it involved prime numbers 🙂

1. Glnc says:

You’re right! I misremembered those two results. Ones a lot more modern than the other even though they sound so similar. But yes, assuming the Green tao theorem then we can just take a sequence of 8 primes in arithmetic progression with each other and make the first 4 your xs, and last 4 your ys. Using these arithmetic progression reduces the problem to labeling the vertices of the cube with 0,…,7 which is much simpler! By green tao there are infinitely many such progressions, so we do get infinitely many solutions. But I find it nice that you found some small magnitude examples where there is less regularity than you get from my solution

2. Eric says:

The dodecahedron is significantly simpler than that, though my method does not give an infinite number of solutions (which would require the prime k-tuple conjecture to be proven). Essentially:

(1) Choose a prime 5-tuple with members $p \geq 7$. You can make it as larges as you desire, but the calculations will probably be easier form smaller starting primes. These primes are $(a,b,c,d,e)$.

(2) Use your favorite programming language to look for parallel tuples, i.e., prime 5-tuples with the same differences, offset by some $n$. Take the first four, and the *offsets* are $(f, g, h, i)$, with $j=0$.

I haven’t done a thorough search, but the smallest solution I’ve found has $(a,b,c,d,e,f,g,h,i) = (11, 31, 41, 61, 71, 12, 42, 96, 348)$, with $419$ being the highest prime. I think it’s likely there are smaller ones, though probably not too many.

(Hopefully I’m right that your page interprets MathML or something similar, or this’ll look really weird.)