What are the orientability of the real projective space $\mathbb{RP}^n$, the complex projective space $\mathbb{CP}^n$ and the quaternionic projective space $\mathbb{HP}^n$, for any $n$?
Are the $\mathbb{RP}^n$, $\mathbb{CP}^n$ and $\mathbb{HP}^n$ spin-manifold or not, for any $n$?
p.s. What is known: I know $\mathbb{RP}^n$ is non-orientable for even $n$ and orientable for odd $n$. I know orientable manifolds for $n \leq 3$ all are spin manifold. I know $\mathbb{CP}^4$ is orientable but not spin manifold.
Partial answers are welcome.
Following the canonical isomorphism $H_k(\Bbb{RP}^n;\Bbb Z/2) = \Bbb Z/2$, it is well-known (see e.g. Milnor and Stasheff) that $w_k(\Bbb{RP}^n) = \binom{n+1}{k} \mod 2$. In particular, $w_1(\Bbb{RP}^n) = n+1$ and thus is zero iff $n$ is odd, and $w_2(\Bbb{RP}^n) = n(n+1)/2 \pmod 2$ for $n \geq 2$, and thus is zero iff $n = 0, 1$, or $n>2$ and $n$ is equal to $n=0$ or $3$ mod $4$.
One has identical results for $\Bbb{CP}^n$ and its Chern classes (which reduce mod 2 to its even Steifel-Whitney classes), and so $w_2(\Bbb{CP}^n)$ is zero iff $n$ is odd, and of course its $w_1$ is always zero as it has trivial first cohomology.
$\Bbb{HP}^n$ has trivial first and second cohomology so its $w_1$ and $w_2$ are automatically zero (and satisfies the same formula above for $w_{4k}$.)
A manifold is orientable iff $w_1(M) = 0$, and is spin iff both $w_1$ and $w_2$ are zero. So $\Bbb{RP}^n$ is orientable for odd $n$, and spin for $n = 4k+3$; $\Bbb{CP}^n$ is always orientable and spin when $n$ is odd; and $\Bbb{HP}^n$ is always spin.