What's true in $\mathbb{R}^4$, false in $\mathbb{R}^3$ and uninteresting in $\mathbb{R}^5$?

Due to the theory of quaternions, due to Hamilton, $\bf R^4$ has a structure of a of non commutative field. The only dimension for which $\bf R^n$ is a field are $n=1,2, 4$ . As an application, the special orthogonal group in dimension 4 is not simple : it is the quotient of $U\times U$ by it center $Z/2Z$ where $U$ is the unitary group in complex dimension 2, or the set of quaternions of norm 1. In other dimension (other than 2) the special orthogonal group modulo its center is simple.

Lets bump knot theory up a dimension. In general, an $n$-sphere can be non-trivially knotted in $\mathbb{R}^{n+2}$. Obviously, an $n$-sphere can be embedded in $\mathbb{R}^{n+1}$, where it is usually defined, but no knotting can occur. In $\mathbb{R}^{n+3}$, we can use the extra dimension and unknot every $n$-sphere.

So, for $n=2$, we have that a $2$-sphere can be knotted in $\mathbb{R}^{4}$, cannot be knotted in $\mathbb{R}^{3}$, and is uninteresting in $\mathbb{R}^{5}$, since there is only one knot type.

If you want to look into this more, you should look at any number of great books: Rolfsen, Adams, and more that I am just not thinking of at the moment.

Edit: As per Mike's comment, we should be assuming PL or locally flat here.

Only for $n=4$ does there exist an open set $U\subseteq\mathbb{R}^n$ that is homeomorphic to $\mathbb{R}^n$ but not diffeomorphic to $\mathbb{R}^n$ (a small exotic $\mathbb{R}^4$). What this means is not too difficult to explain (no need to explain what a manifold is, only what a homeomorphism and a diffeomorphism are between open subsets of $\mathbb{R}^n$). I don't think it qualifies as "uninteresting for $\mathbb{R}^5$", though (it's definitely not a triviality in any dimension other than $1$), but you seemed to say "false" was also OK.