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.




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**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...



2-dimensional sublattices with all vectors having very big square (in absolute value)
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.


