2-dimensional sublattices with all vectors having very big square (in absolute value) 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...
 A: This doesn't solve the question : it shows that a non-degenerate lattice always contains a primitive rank 2 sublattice with the required property regarding norms ... but the sublattice found is definite. Maybe a modification of the argument would yield the desired result, so I post it.
Without loss of generality we may assume $\Lambda$ is a $3$-dimensional maximal lattice. 
Let us write $B_N=\{v\in\Lambda, (v,v)\leq N\}$.
The positive definite case seems easy : for a given $N$, the ball $B_N$ in $\Lambda$ contains only a finite number of vectors, and there certainly exists a $2$-dimensional lattice that doesn't intersect $B_N-\{0\}$. 
The lorentzian case is a little subtler since then $B_N$ is not finite anymore. Here is an argument : let us see $\Lambda$ as a lattice in the standard Lorentzian space $\mathbf R^3$ with the form $q(x,y,z)=-x^2+y^2+z^2$ and let us choose the embedding so that the intersection of $\Lambda-\{0\}$ and the hyperplane $P$ defined by the equation $x=0$ is empty. The idea is that the set of vectors of $B_N$ that lie near $P$ is finite, while there are infinitely many hyperplanes of $\Lambda$ that are as near as you want from $P$. 
More precisely : let $x_N$ be the minimum of $\vert x\vert $ on $B_N-\{0\}$ (the chosen embedding implies $x_N>0$),  let $B_N'$ be the finite set of vectors of the form $(\pm x_N,y,z)$ in $B_N$ . Let $y_N$ (resp. $z_N$) be the maximum of $y$ (resp. $z$) on $B'_N$. 
Finally let $C$ be the set of vectors $(a,b,c)$ satisfying $\vert b\vert \leq \vert \frac{x_N}{3y_N} a\vert $ and $\vert c\vert\leq \vert \frac{x_N}{3z_N} a\vert$. This is a cone with a non-empty interior in the lorentzian space. Thus its intersection with $\Lambda-\{0\}$ is non-empty. Let us pick a vector $v=(a,b,c)$ therein, with $a\neq 0$. Then for a vector $(x,y,z)$ of the ortogonal complement of $v$, the inequalities $\vert y\vert\leq y_N$ and $\vert z\vert\leq z_N$ imply the inequality $x<x_n$. Thus we have $v^\perp\cap B_N=\{0\}$, and $v^\perp\cap\Lambda$ is a solution.
A: 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.
A: I am grateful to Ekaterina Amerik for this remark.
This is not an answer, but in a special case (not the one I need) the solution is known.
In David Morrison's paper 
"On K3 surfaces with large Picard number", Corollary 2.5, it is shown that any
non-degenerate even lattice of appropriate signature and rank admits a primitive embedding to a given unimodular even lattice (I suppose that this is also true for odd lattices, though it's hard to find in the literature). Then we need to take a lattice $N\cdot H$ obtained from any given lattice of signature (1,1) by rescaling with a factor N, and it would be embeddable to a unimodular lattice, solving the problem for unimodular lattices. 
Unfortunately, the applications that we have need lattices which are not unimodular.
