Solution: Let $ x = \{\sqrt {a}\}$, and notwithstanding the statement, assume that $ S_n$ is a rational number for some natural $ n$. Due to the well-known identity: \[ x^{n + 1} - 1 = (x - 1)\cdot S_n \] Clearly, the number $ x^{n + 1}$ is expressible in the form $ m + l\sqrt {a}$, for some $ m,l\in\mathbb{Z}$. (This is because $ x = \sqrt {a} - [\sqrt {a}]$). Therefore, \[ m + l\sqrt {a} = x\cdot S_n - S_n = \sqrt {a}\cdot S_n - [\sqrt {a}]S_n - S_n \] ,or \[ \sqrt{a} = \frac {m + S_n([\sqrt {a}] + 1)}{S_n - l} \] contradiction.

I don't see contradiction, what if we take $ l=S_n$?

Actually $ l > S_n$, but well, I should have explained it better: Note that $ a\geq 2$. Clearly, $ S_n < n$. If $ n = 2k$, the coefficient in front of $ \sqrt {a}$ in $ x^{2k + 1} = (\sqrt {a} - [\sqrt {a}])^{2k + 1}$ is $ (a^{k} + \binom{2k + 1}{2}[\sqrt {a}]^2\cdot a^{k - 1} + \dots ) > (2k + 1)k > (n + 1)$ the case $ n = 2k + 1$ is totally similar. Hopefully, I haven't made a mistake.