Euler number of K3 surface The Euler number of a K3 surface is 24, which can be obtained by using many deep results in algebraic geometry.  Is there an elementary way in algebraic topology to get it? For example, let's consider $X=\{z_{0}^{4}+ \cdots +z_{3}^{4}=0\} \subset \mathbb{CP}^{3}$. Is it possible to cut $X$ into pieces, and then use Mayer-Vieroris sequence?  Or some simple Morse function will do the job?  
P.S. First of all, thank everybody for answers and comments. As a matter of fact, I'm searching a proof within the reach of standard algebraic or differential topology text book. I can accept that all smooth hyperface K3 are  diffeomorphic to each other, which may be obtained by applying Sad's theorem (or like some arguments in the Morse theorem). However R-R or Torelli are too much. I like to consider the family $X_{t}=\{t(z_{0}^{4}+ \cdots +z_{3}^{4})+z_{0}^{4}....z_{3}^{4}=0\}$, and let $t$ tend to $0$. The limit is the union of several hyperplanes. I want to construct a decomposition when $t$ close to $0$, in which every piece diffoemorphic to a subset of the hyperplane or the normal crossing singularities.  Then we can apply Mayer-Vieroris argument. However it is really messy, and I stop to think this question. 
 A: The given $K3$ surface $X \subset \mathbb P^3$ of degree 4 is defined by the exaxt sequence
$$0 \longrightarrow \mathscr O_ {\mathbb P^3}(-4) \longrightarrow \mathscr O_ \mathbb {P^3} \longrightarrow \mathscr O_X \longrightarrow 0.$$
Hence the canonical bundle $\kappa_X \cong \mathscr O_X(-4+4) = \mathscr O_X$, which implies $c_1^2(X) = 0$.
The holomorphic Euler characteristic is additive on exact sequences. Hence
$$\chi(\mathscr O_X) = \chi (\mathscr O_\mathbb {P^3}) -\chi (\mathscr O_\mathbb {P^3}(-4)).$$
The cohomology of $\mathscr O_{\mathbb P^3}(k)$ is well-known. We obtain $\chi(\mathscr O_X) = 1 - (-1) = 2$.
Eventually, we invoke the Noether formula 
$$\chi(\mathscr O_X) = \frac {c_1^2(X) + c_2(X)} {12}$$ 
obtaining the topological Euler number $\chi_{top}(X) = c_2(X) = 2*12 = 24$. 
Apparently, also this proof uses some non-trivial results from algebraic geometry, e.g., the Noether formula. But if you strive for the general statement about the diffeomorphism type of K3 surfaces, you have to invoke deformation theory of K3 surfaces with even deeper results than Simon's or my proof above. (See Barth, W.; Hulek, K.; Peters, Ch; van de Ven, A.: Compact Complex Surfaces. VIII, Cor. 8.6) 
A: One way you could look at it (although how deep this is depends on your perspective, I guess) is as follows.
Consider to begin with a complex torus $\mathbb{C}^2 / \Lambda$. This has a natural involution given by multiplication by $\pm1$, and yields a singular quotient. You can check easily that the dimensions of the cohomology groups of the quotient are
$$
h^0 = 1 \qquad h^2 = 6 \qquad h^4 = 1 \qquad h^3 = h^1 = 0
$$
Now, this quotient has 16 singular points, all of which locally look like $\mathbb{C}^2/\pm1$. These can be resolved by blowing up once to yield a singular exceptional $\mathbb{P}^1$, and so each of these increase the Euler number of the resolution by 1. The resulting surface is a K3 surface, and has Euler number $1 + 6 + 1 + 16 = 24$ as desired.
Now given that K3 surfaces are all diffeomorphic...
Again, how much depth this has may depend on your perspective. How comfortable are you with cohomology of global quotients in terms of the original? How about computing resolutions of singular surfaces? And finally, how comfortable are you with the fact that all K3 surfaces are diffeomorphic?
A: You can avoid using Noether's formula by computing the Chern classes of a quartic surface in $\mathbb{CP}^3$ from the exact sequence in jo wehler's answer. I do this in this blog post; the computation is not hard and generalizes to give the Euler characteristic of any complete intersection. I don't know what you mean by "many deep results" but this computation doesn't even use anything as hard as Riemann-Roch, as far as I can tell. 
I think a more direct topological argument using a generalization of the Riemann-Hurwitz formula is also possible, but I tried writing it down once and it got surprisingly messy. 
