MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The $4$-dimensional lattice $\mathbb{Z}^{4}$ has vectors of length $\sqrt{n}$ for any positive integer $n$ by the Four Squares Theorem, but this need not be true for higher-dimensional integral, unimodular lattices for two reasons:

(i) Some small positive integers could be skipped as squared lengths of lattice vectors. For example, the odd Leech lattice has no $v$ with $v \cdot v = 1$ or $2$.
(ii) The lattice may be even.

Therefore, the way to word the question to recognize these possibilities is:

Let $\Lambda$ be an integral unimodular lattice of dimension $d$, where $d \geq 4$.
(i) If $\Lambda$ is odd, then is it true that every sufficiently large positive integer arises as the squared length of a vector in $\Lambda$?
(ii) If $\Lambda$ is even, then is it true that every sufficiently large even positive integer arises as the squared length of a vector in $\Lambda$?

share|cite|improve this question

Yes, both are true. For example, see Theorem 1.6 of Chapter 11 of Cassels's "Rational Quadratic Forms", which says that if $Q$ is a positive definite integral quadratic form, then there is an integer $N$ depending on $Q$ such that if $a > N$ and $a$ is represented primitively by $f$ over all $\mathbb{Z}_p$ then $a$ is represented by $Q$. The local primitive representability is easy to show using the fact that $f$ is unimodular, and the classification of forms over $\mathbb{Z}_p$ by invariants.

For instance, if $p$ is odd and $p \nmid a$, then $f$ is equivalent over $\mathbb{Z}_p$ to $(a, a \det(f), 1, 1, \dots, 1)$, which obviously represents $a$ primitively. If $p | a$ you could use $((a-1), (a-1) \det(f), 1, 1, \dots, 1)$ which represents $a$ primitively. I won't do the analysis for $p = 2$, but see section 4 of chapter $8$ of Cassels.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.