Which numbers are sums of finite numbers of reciprocal squares?

Question

Question: Is there a “nice” characterization of rational numbers $q$ for which $q$ can be written as

$$q = \frac{1}{n_1^2} + \frac{1}{n_2^2} + \dots + \frac{1}{n_k^2}$$

for distinct natural numbers $n_1 \lt n_2 \lt \dots \lt n_k$?

I’m aware of several related results, but none of them seem applicable here:

• Lagrange’s four-square theorem says that every natural number is the sum of four squares, but that doesn’t transfer to reciprocals of squares.
• The Basel problem states that $\sum_{n=1}^\infty{1 \over n^2} = {\pi^2 \over 6}$, which gives an upper bound on the numbers that can be written this way. If we allow for infinitely many terms in the sum I think (?) this would mean we can express every number up to $\pi^2 \over 6$ this way, but I’m interested only in finite sums.
• The existence of Egyptian fraction representations and the divergence of the harmonic series ensures every nonnegative rational number can be written as a sum of distinct unit fractions, but those fractions aren’t necessarily constrained to be inverse squares.

Is there a simple characterization of the rational numbers that can be written this way?

Answer 1

You can't express every rational number up to $\pi^2/6$. For example, $3/4$ is impossible - the sum can't contain 1, but then it's bounded above by $\pi^2/6 - 1$, which is less than $3/4$. However this sort of argument only seems to exclude the interval $[\pi^2/6 - 1, 1)$.

And it turns out (much less obviously) that that's the only obstruction - see R. L. Graham, On Finite Sums of Reciprocals of Distinct nth powers, Pacific Journal of Mathematics, 1964. In particular Corollary 1 says a rational $p/q$ is a sum of distinct reciprocals of squares if and only if $$p/q \in [0, \pi^2/6-1) \cup [1, \pi^2/6).$$

Graham wrote that this was an unpublished result of Erdos. Graham also wrote at the end of the paper that "It is not difficult to obtain representations of specific rationals" as such sums, but he gave no hint of how he did it. The proof appears to be "elementary" in the sense of not using any heavy machinery.