QUESTION: Let $\Lambda\times\Lambda\rightarrow {\Bbb Z}$ be a lattice, that is, ${\Bbb Z}^n$ with a non-degenerate integer quadratic form, not definite, not necessarily unimodular, $n>2$. I want to show that for each $N>0$ there exists a 2-dimensional primitive sublattice $\Lambda_0\subset \Lambda$, also not definite, such that all nonzero vectors $x\in \Lambda_0$ satisfy $|(x,x)| > N$.

This question is motivated by considering a K3 surface (or a hyperkahler manifold) with 2-dimensional Picard lattice. It is known that the squares of minimal rational curves are bounded, and we want to find a manifold with Picard number 2 having no minimal rational curves.

Bounty added This problem has kept me up late the last two nights, so I'm hoping some of the quadratic form experts will work on it.

I tried to find a strategy working with rational quadratic forms rather than integers, but failed for the following reason: Equip $\mathbb{Q}^3$ with the quadratic form $x^2+y^2-z^2$. Then I claim that any rank two subspace $L$ on which this form is non-degenerate contains a vector of norm $1$.

Proof: Let $L^{\perp} = \mathbb{Q} v$. Since our form is nondegenerate on $L$, we have $\langle v,v \rangle \neq 0$, say $\langle v,v \rangle = N$, and $\mathbb{Q}^3 = L \oplus \mathbb{Q} v$. Now, $x^2+y^2-z^2$ is equivalent to $N (x')^2 + (y')^2 - N (z')^2$, by the change of variables $(x,y,z) = (\tfrac{N+1}{2} x' + \tfrac{N-1}{2} z', y', \tfrac{N-1}{2} x' + \tfrac{N+1}{2} z')$. So our form is equivalent to $L' \oplus \mathbb{Q} v$, where the form on $L'$ is $(y')^2 - N (z')^2$. By Witt cancellation, the forms on $L$ and $L'$ are equivalent. Since $(y')^2 - N (z')^2$ represents $1$, so does our form on $L$. $\square$

So, when we are trying to construct rank two sub-lattices with no vectors of norm $1$, we have to do so using lattices which do represent $1$ rationally. This seems hard to me...