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...