Let $X$ be a projective variety over $\Bbb C$. One can compute the integral cohomology groups $H^p(X,\Bbb Z)$ by looking at the constant sheaf $\Bbb Z$ in the Zariski topology on $X$, but one can do the same with respect the Euclidean topology on $X$ as well.
Question: what guarantees that the two definitions will give the same cohomology groups? My initial reaction was that this should follow from GAGA, but $\Bbb Z$ is not a coherent $\mathcal O_X$-module, so not sure what to do.
I am sure this is addressed in the literature, so a reference would be fine as well.
These are not at all the same. If $X$ is irreducible, then $H^q_{Zar}(X, \mathbb{Z})$ will be $0$ for $q>0$.
Proof: If $U_1$, $U_2$, ..., $U_n$ are nonempty Zariski open subsets of $X$, then $\bigcap_{i=1}^n U_i$ will be nonempty and connected. So the Cech complex of $\mathbb{Z}$ with respect to any cover with $N$ open sets will be $\mathbb{Z}^{\binom{N}{k+1}}$ in degree $k$. This is the complex which computes the cohomology of an $N$-1 dimensional simplex, and is thus acyclic.
Alternate (probably better) proof: If $X$ is irreducible and $U \subseteq V$ are nonempty open sets, then $U$ and $V$ are both connected, so the restriction map $\underline{\mathbb{Z}}(U) \longrightarrow \underline{\mathbb{Z}}(V)$ is an isomorphism, and thus the sheaf $\underline{\mathbb{Z}}$ is flasque.
If you want to do something algebraic and get the classical betti numbers as an answer, you either have to use something like etale cohomology or else something like the de Rham complex.
EDIT TO ADD: If you want to get cohomology with $\mathbb{Z}$ coefficients as an answer, you honestly have to work analytically, no algebraic method will work. Consider elliptic curves with complex multiplication by $\mathbb{Z}[\sqrt{-5}]$. Over $\mathbb{C}$, there are two isomorphism classes, with $j$-invariants $-632000 \pm 282880 \sqrt{5}$. Let $E_+$ and $E_-$ be the ones with $+$ and $-$ signs. We have $H^1(E_+, \mathbb{Z}) \cong H^1(E_-, \mathbb{Z}) \cong \mathbb{Z}^2$ as abelian groups. However, the complex multiplication structure makes them into modules for $\mathbb{Z}[\sqrt{-5}]$ and they are not isomorphic as such: $H^1(E_-, \mathbb{Z})$ is free of rank $1$, where as $H^1(E_+, \mathbb{Z})$ is isomorphic to the nonprincipal ideal $\langle 2, 1+\sqrt{-5} \rangle$.
Why is this bad news? We can choose a model $\mathcal{E}$ for $E_+$ over $\mathbb{Q}(\sqrt{5})$, with complex multiplication $\phi$ given by a morphism defined over $\mathbb{Q}(\sqrt{5}, \sqrt{-5})$. Let $\sigma$ be a Galois symmetry of $\mathbb{Q}(\sqrt{5}, \sqrt{-5})$ with $\sigma(\sqrt{5}) = - \sqrt{5}$. Then $\sigma(\phi)$ is an endomorphism of $\sigma(\mathcal{E})$, and $\sigma(E_+)$ is $E_-$. If we had an algebraic way of describing the functor $H^1(\ \ , \mathbb{Z})$, then we would have an isomorphism between $H^1(\mathcal{E}, \mathbb{Z})$ and $H^1(\sigma(\mathcal{E}), \mathbb{Z})$, conjugating $\phi^{\ast}$ to $\sigma(\phi^{\ast})$. (Because everything algebraic commutes with Galois symmetries.) But $\sigma(\mathcal{E})$ is $E_-$ as a complex manifold, so this contradicts that $H^1(E_-, \mathbb{Z})$ and $H^1(E_+, \mathbb{Z})$ are not isomorphic as $\mathbb{Z}[\sqrt{-5}]$ modules.