Why are There No "Triernions" (3-dimensional analogue of complex numbers / quaternions)?
It's because there isn't one! (Indeed, Hamilton was originally searching for such a thing, and found the quaternions instead; it was only later that people understood why he hadn't been successful, initially.)
The quaternions - along with the real numbers and the complex numbers - have a number of nice properties: specifically, they form a real division algebra. This is a mouthful, but basically amounts to:
Addition/multiplication of quaternions satisfy the ring axioms.
We can divide by quaternions.
We can multiply a quaternion by a real (and this "scalar multiplication" satisfies the basic properties it should).
It turns out the only finite-dimensional real division algebras are $\mathbb{R}$, $\mathbb{C}$, and the quaternions; see this. (I include associativity in the definition of algebra: if we allow non-associative algebras, then the octonions also count.)
By the way, there is a way to (sort of) keep going past the quaternions: the Cayley-Dickson construction. This produces things like the octonions and the sedenions, and other delightfully weird algebraic structures. However, it has a couple drawbacks:
Each time you apply Cayley-Dickson, the dimension of the starting algebra doubles. So this won't help us get to $3$.
Also, you keep losing nice properties. Passing from the reals to the complex numbers, we lose order; going from the complexes to the quaternions, we lose commutativity of multiplication. If we keep going, we lose associativity of multiplication, in increasing degrees: the sedenions are even less associative than the octonions, etc.
Assume $A$ is a three-dimensional (associative) algebra over $\mathbb{R}$. We can assume $\mathbb{R}$ is embedded in $A$. If $a\in A$, $a\notin\mathbb{R}$ the map $l_a\colon A\to A$, $l_a(x)=ax$, is an endomorphism of $A$ as a vector space over $\mathbb{R}$.
Let $\lambda$ be a real eigenvalue of $l_a$, with eigenvector $b\ne0$, so $ab=\lambda b$. Such an eigenvalue exists, because the characteristic polynomial of $l_a$ has degree $3$. Then $(a-\lambda)b=0$. Note that $a-\lambda\ne0$, so $A$ has zero divisors, in particular $A$ is not a division algebra.
It's a bit more complicated showing that a finite-dimensional division algebra over $\mathbb{R}$ can only have dimension $1$, $2$ or $4$ and it is isomorphic to $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$ (the quaternions); this is known as Frobenius' theorem.
On the other hand, three-dimensional algebras over $\mathbb{R}$ exist (but they have zero divisors, as shown above). A simple example is $\mathbb{R}[X]/(X^3-1)$, but they can be non-commutative as well.
There is an algebra in dimension 6, halfway between 4 and 8, that is not a division algebra but correctly interpolates various constructions.
http://arxiv.org/abs/math/0411428
Nothing like that is known for dimension 3 sitting between 2 and 4.