nicetry007 wrote: Suppose $ n = k^{2}+c$ where $ 0 \;< \;c\;<2k+1$. Then $ \lfloor \sqrt{n}\rfloor = k$. Suppose $ gcd(k,c) = d$ and $ k = d\cdot k_{1}$ and $ c = d\cdot c_{1}$. We have $ k^{3}\;|\; (k^{2}+c)^{2}\;\Leftrightarrow \;d^{3}k_{1}^{3}\;|\; d^{2}\;(dk_{1}^{2}+c_{1})^{2}\;\Leftrightarrow \;dk_{1}^{3}\;|\; d^{2}k_{1}^{4}+2dc_{1}k_{1}^{2}+c_{1}^{2}\;$ $ \Rightarrow \; k_{1}^{2}\;|\; d^{2}k_{1}^{4}+2dc_{1}k_{1}^{2}+c_{1}^{2}\;\Rightarrow \;k_{1}^{2}\;|\; c_{1}^{2}\;\Rightarrow \;k_{1}\;|\; c_{1}$. But $ gcd(k_{1},c_{1}) = 1$. This implies $ k_{1}= 1$ and $ d = k$. Hence, $ n = k^{2}+c_{1}k$ and $ c_{1}= 1$ or $ 2$. Plugging $ k_{1}= 1$, $ d = k$ and $ c_{1}= 1$ in $ dk_{1}^{3}\;|\; d^{2}k_{1}^{4}+2dc_{1}k_{1}^{2}+c_{1}^{2}$, we get $ k\;|\; k^{2}+2k+1\; \Rightarrow \;k = 1\; \Rightarrow \;n = 2$. Plugging $ k_{1}= 1$, $ d = k$ and $ c_{1}= 2$ in $ dk_{1}^{3}\;|\; d^{2}k_{1}^{4}+2dc_{1}k_{1}^{2}+c_{1}^{2}$, we get $ k\;|\; k^{2}+4k+4\; \Rightarrow \;k = 1,\;2$ or $ 4 \; \Rightarrow \;n = 3,\;8$ or $ 24$. Hence, $ n = 2,\; 3,\;8,\;24$.