If n is integer from $ a=[n]$ we get $ a=n$. Therefore from $ a^{3}|n^{2}$ we get $ n=\pm 1$.

I think it does not have sense if the test is correct. The problem may have sense if it is $ a=[\sqrt{n}]$, in this case i think the answer is $ n=24$

salva90 wrote: I think it does not have sense if the test is correct. The problem may have sense if it is $ a = [\sqrt {n}]$, in this case i think the answer is $ n = 24$ I think so .I will find all solution of n. Let $ n = p^2 + r$ where $ 0\leq r < 2p + 1$ $ \Longrightarrow [\sqrt {n}] = p$ Imply that : $ p^3|(p^2 + r)^2$ (1) So $ p|r,\Longrightarrow r\in \{0,p,2p\}$ Case 1 $ r = 0$ Easy to check it is true for all p . So $ n = p^2$ Case 2 $ r = p$ (1)$ \Longleftrightarrow p^3|(p^2 + p)^2$ $ \Longleftrightarrow p|(p + 1)^2$ $ \Longleftrightarrow p = 1$ So $ n\in \{1,2\}$ Case 3 $ r = 2p$ (1)$ \Longleftrightarrow p|(p + 2)^2$ So $ p|4$ So $ n\in \{3,8,15,24\}$ So solution is $ n\in \{t^2,1,2,3,8,15,24\}$

Yes, I've got the same solution But be careful: you said that $ n = t^2$ is a solution, instead the conditions say n cannot be a perfect square

2楼说的是对的，原题好像是

2楼说的是对的，原题好像是 a=[\sqrt{n}]

Yes, I've got the same solution But be careful: you said that n = t^2 is a solution, instead the conditions say n cannot be a perfect square and n=15alsoconnotbeaanswer