Jose writes: "if you have a polynomial that is positive semidefinite, more likely than not it is a sum of squares"
This sort of statement is very sensitive to what probability distribution you use on the space of all polynomials. I am sure there are formulations in which it is true; here is one in which it is false.
Fix d greater than 1. Let Poly(2d,n) be the vector space of homogenous*, degree 2d polynomials in n variables. Let's choose polynomials uniformly at random from unit sphere in this space. Blekherman has computed that the probability that a polynomial is positive is ~ n^{-1/2}, while the probability that it is a sum of squares is ~ n^{-d/2}. So, for n large, almost all positive polynomials are not sums of squares.
Blekherman also has a recent preprint showing that, in the same sense, almost all positive convex polynomials are not sums of squares.
* If you don't like working with homogenous polynomials, notice that Poly(2d,n) is also the vector space of inhomogenous polynomials in n-1 variables with degree at most d. Just plug in 1 for the last variable. Under this correspondence, a polynomial is nonnegative on R^n if and only if it is nonnegative on R^{n-1} x {1}. The property of being strictly positive
is not preserved by this transformation, but the polynomials which are nonnegative and not strictly positive form a set of measure 0.
Let $K$ be a complex quadratic number field such that $K(i)$ has class number $1$. If $K$ has class number $\ne 1$, then $K(i)$ must be the Hilbert class field of $K$, which, in this case, coincides with the genus class field of $K$. By genus theory, the discriminant of $K$ must have the form $d = -4p$ for a prime number $p \equiv 1 \bmod 4$.
In these cases, the ring of integers in $K$ is $R = {\mathbb Z}[\sqrt{-p}]$, and the ring of integers in $L$ is ${\mathbb Z}[i, (1 + \sqrt{p})/2] \ne R[i]$.
Finding number fields of higher degree with this property seems to be an interesting problem;
I can't think of an obvious approach in general, but if I find anything, I'll let you know.
Edit 1. The argument works for all imaginary number fields: if $R$ is the ring of integers in a number field $K$, and if $S = R[i]$ is the ring of integers in the extension $L = K(i)$, then disc$(L) = \pm 4$ disc$(K)^2$ (this is a simple determinant calculation: take an integral basis
$\{\alpha_1, \ldots, \alpha_n\}$ for $K$; then $\{\alpha_1, \ldots, \alpha_n, i\alpha_1, \ldots, i\alpha_n\}$ is an integral basis for $L$).
On the other hand, if $L$ has class number $1$ and $K$ is not a PID, then $K$ has class number $2$ and $L$ is the Hilbert class field of $K$. This implies disc$(L) = \pm$ disc$(K)^2$.
The problem in the non-imaginary case is that $K(i)$ might be unramified at all finite primes, but not at infinity; in this case, $K$ has class number $2$ in the strict sense, yet its ring of integers is a UFD.
Remark 2. By looking at $p = \alpha^2 + \beta^2$ modulo $4$ it follows (unless I did something stupid) that primes $p \equiv 3, 7 \bmod 20$ are not sums of two squares in
${\mathbb Z}[\sqrt{-5}]$.
Edit 2. Your suggestion to look at rings $R[\frac12]$, where $R$ is the ring of integers of a quadratic number field, seems to work for $K = {\mathbb Q}(\sqrt{-17})$, which has a cyclic class group of order $4$. The ring $R[\frac12]$ has class number $2$ because $2$ is ramified and so generates a class of order $2$, and $S = R[\frac12,i]$ is the integral closure of $R[\frac12]$ in the extension $L = K(i)$. The ring $R[\frac12]$ has class number $2$, and $S$ is a UFD since $L$ has class number $2$, and its class group is generated by one of the prime ideals above $2$ (the ramified prime above $2$ in $K$ splits in $L$).
Edit 3. The corollary concerning sums of two squares in $R[\frac12]$ shows that primes $p \equiv 3 \bmod 4$ splitting in $K$ are sums of two squares up to units. In fact it follows from genus theory that the prime ideals above such $p$ are not in the principal genus, hence lie in the same class as the prime above $2$ or in its inverse. This implies that either $2p = x^2 + 17y^2$ or $8p = x^2 + 17y^2$, giving a representation of $p$ as a sum of two squares up to a unit (remember $2$ is invertible). Thus everything is working fine.
Best Answer
The answer is similar to one provided here. The approach is elementary and proves stronger fact that it can be expressed as sum of two coprime squares.
We consider the product $(n+2)(n^3-2)$ which is equal to $n^{4}-2n+2n^3-4$. Now we observe that $$(n+2)(n^3-2)=(n^2+n-7)^2+(13n^2+12n-53).$$ Proceeding as shown in the link we can easily get that $13n^2+12n-53$ is a perfect square for infinitely many $n\equiv 3\pmod{4}$ with first solution as $13(3^2)+12(3)-53=10^2$. If one chooses such a $n$ then clearly $n^3-2$ doesn't have any prime divisor of form $4k+3$ as $\gcd(n^2+n-7,13n^2+12n-53)\mid 25\cdot 59$ and $59$ doesn't divide $n^3-2$ if one looks at the general solution of the equation. Since, $n^3-2$ doesn't have any prime divisor of form $4k+3$ and $n\equiv 3\pmod{4}$ we can infer from the folklore result that it can be expressed as sum of two coprime squares.