= Solution
For a complex <K3 surface>, $L=H^2(X,\mathbb Z)$ is torsion-free and has its integral <cup product> pairing
$$
q(x,y)=\langle x\smile y,[X]\rangle.
$$
The <K3 intersection lattice> is
$$
\boxed{L\cong U^{\oplus3}\oplus E_8(-1)^{\oplus2},}
$$
the even unimodular <integral quadratic lattice> of signature $(3,19)$. Here $U$ is the <hyperbolic plane (quadratic form)> with Gram matrix
$$
\begin{pmatrix}0&1\\1&0\end{pmatrix},
$$
and $E_8(-1)$ is the negative of the positive-definite <E8 lattice>. Thus the rank is $3\cdot2+2\cdot8=22$ and the pairing has determinant of absolute value one.
These invariants can be checked from the topology and <Hodge theory>. The trivial <canonical bundle> and $H^1(X,\mathcal O_X)=0$ give $\chi(\mathcal O_X)=2$. The <Noether formula> gives $c_2(X)=24$, which is the topological <Euler characteristic>. Since $X$ is simply connected, $b_1=b_3=0$, so $24=2+b_2$ and $b_2=22$. Integral <Poincare duality> gives unimodularity; the universal coefficient sequence and $H_1(X,\mathbb Z)=0$ give torsion-freeness of $H^2(X,\mathbb Z)$.
Its second <Stiefel–Whitney class> is the reduction modulo two of $c_1(X)$, hence zero. The characteristic-class identity
$$
\langle x^2,[X]\rangle
\equiv\langle x\smile w_2(X),[X]\rangle\pmod2
$$
makes the lattice even. Finally the real and imaginary parts of a holomorphic two-form give two positive directions, a <Kähler class> gives a third, and the primitive real $(1,1)$ classes are negative definite by the <Hodge–Riemann bilinear relations>. Therefore the signature is $(3,19)$. The classification of indefinite even unimodular lattices then identifies the displayed lattice uniquely. In particular the <Picard group> is only the integral $(1,1)$ part; it should not be confused with this rank-twenty-two topological lattice.
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