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Suppose, on the contrary, that the function $f$ is periodic with the period $T$ for some integer $n\ge 2$. Hence, $f(T)=f(0)=n$. Now we have $$f(T)=\cos (T\sqrt{1})+\cos (T\sqrt{2})+\cdots +\cos (T\sqrt{n})=n,$$ from which we conclude that $\cos (T\sqrt{1})=\cos (T\sqrt{2})=\cdots =\cos (T\sqrt{n})=1$. So, $T=2k\pi$ and $T\sqrt{2}=2l\pi$, where $k,l\in \mathbb{N}$. Hence, $\sqrt{2}=l/k\in \mathbb{Q}$, which is contradiction. In conclusion, there is no integer $n\ge 2$ such that $f$ is periodic.