TITLE: 2-dimensional sublattices with all vectors having very big square (in absolute value)
QUESTION [12 upvotes]: 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...

REPLY [3 votes]: For the record: OP Misha Verbitsky writes that
"[$\Lambda$ of] rank $\geq 6$ or $\geq 7$ is a usual assumption
in these kind of applications", in which case 
Ekaterina Amerik's suggestion of using
$N\cdot H$ (for some fixed indefinite rank-$2$ form $H$)
surely yields the easiest construction;
but there are also primitive indefinite forms $\Lambda_0$ of rank $2$ 
that represent no small nonzero integers of either sign,
and those should work even for $\Lambda$ of rank as small as $3$ 
(as in the lattice that troubled the OP in the first place).
One reasonably simple construction is to take $M>N$ odd and set 
$$
Q(x,y) = \frac1M ((x+My)^2 - (M^4+1)x^2)
= -M^3 x^2 + 2xy + M y^2,
$$
which represents $\pm M$ (as for $(x,y)=(0,1)$ and $(M,M^2-1)$)
but no integer of smaller absolute value other than zero.
This can be checked using the fact that the associated
real quadratic field ${\bf Q}(\sqrt{M^4+1})$ has fundamental unit
$M^2 + \sqrt{M^4+1}$ (of norm $-1$) as small as possible.
Moreover, unlike the forms $N \cdot H$, this $Q$ is primitive,
so any embedding of the into $\Lambda$ of the indefinite lattice 
$\Lambda_0$ associated to $Q$ is automatically primitive as well.