Let $\mathbb{R}P^n$, $\mathbb{C}P^n$, $\mathbb{H}P^n$ be the real, complex, quaternionic projective spaces resp.
I want to find all $n$ such that $\mathbb{R}P^n$ can be embedded into $\mathbb{R}^{n+1}$.
And find all $n$ such that $\mathbb{C}P^n$ can be embedded into $\mathbb{R}^{2n+1}$.
And find all $n$ such that $\mathbb{H}P^n$ can be embedded into $\mathbb{R}^{4n+1}$.
What is the answer?
A necessary condition for a closed $n$-manifold $M$ to (smoothly) embed or immerse into $\mathbb{R}^{n+1}$ is that its tangent bundle $T$ becomes trivial after adding a single line bundle $L$ (namely the normal bundle of the embedding). This condition is sufficient for an immersion by Hirsch-Smale theory, but the question of embedding is more delicate.
The condition that $T \oplus L$ is trivial implies that $w(T) w(L) = 1$. When $M = \mathbb{RP}^n$, recall that the total Stiefel-Whitney class of $\mathbb{RP}^n$ is $(1 + \alpha)^{n+1}$ where $\alpha \in H^1(\mathbb{RP}^n, \mathbb{F}_2)$ is a generator. If $w_1(L) = 0$ then we need $(1 + \alpha)^{n+1} = 1$, or equivalently ${n+1 \choose k}$ even for $1 \le k \le n$. By Kummer's theorem this happens iff $n+1$ is a power of $2$, so the only candidates are $\mathbb{RP}^{2^k - 1}$. Similarly, if $w_1(L) = \alpha$ then we need $(1 + \alpha)^{n+2} = 1$, and this happens iff $n+2$ is a power of $2$, so the only candidates are $\mathbb{RP}^{2^k - 2}$.
This table of embeddings and immersions of real projective spaces shows that that $\mathbb{RP}^3$ and $\mathbb{RP}^7$ don't embed into $\mathbb{R}^4$ and $\mathbb{R}^8$ respectively, and a 1963 theorem of James rules out $\mathbb{RP}^{2^k - 1}$ for sufficiently large $k$ (I think $k \ge 4$ but I haven't checked). Also, $\mathbb{RP}^6$ doesn't embed into $\mathbb{R}^8$ and $\mathbb{RP}^{14}$ doesn't immerse into $\mathbb{R}^{21}$, so prospects seem poor for this case as well.
For $\mathbb{CP}^n$ we can instead compute the total Pontryagin class, which is $(1 + \alpha^2)^{n+1}$, where $\alpha \in H^2(\mathbb{CP}^n, \mathbb{Z})$ is a generator. The condition that $T \oplus L$ is trivial implies that this vanishes, which is only possible for $n = 1$: for $n \ge 2$ the first Pontryagin class is $(n+1) \alpha^2 \in H^4(\mathbb{CP}^n, \mathbb{Z})$ which doesn't vanish. Hence for $n \ge 2$ we find that $\mathbb{CP}^n$ cannot immerse into $\mathbb{R}^{2n+1}$; this argument in fact shows that the smallest possible codimension of an immersion is $4$ (because this is the smallest codimension for which the normal bundle can have a nontrivial Pontryagin class). Presumably a similar computation works for the total Pontryagin class of $\mathbb{HP}^n$ but I'm less familiar with this case.