Prove that for any prime $p>3$, $\dfrac{p^6-7}{3}+2p^2$ can be written as sum of two perfect cubes.
Proposed by Titu Andreescu.
Solution:
We have
$$\begin{array}{lll} \dfrac{p^6-7}{3}+2p^2 & = & \dfrac{p^6+6p^2-7}{3}\\ & = & \dfrac{9p^6+54p^2-63}{27}\\
& = & \dfrac{(p^6-12p^4+48p^2-64)+(8p^6+12p^4+6p^2+1)}{27} \\ & = & \left(\dfrac{p^2-4}{3}\right)^3 + \left(\dfrac{2p^2+1}{3}\right)^3\end{array}$$ and it is clear that the two summands on the right hand side are integers, since if $p > 3$, $p^2 \equiv 1 \pmod 3$.
$$\begin{array}{lll} \dfrac{p^6-7}{3}+2p^2 & = & \dfrac{p^6+6p^2-7}{3}\\ & = & \dfrac{9p^6+54p^2-63}{27}\\
& = & \dfrac{(p^6-12p^4+48p^2-64)+(8p^6+12p^4+6p^2+1)}{27} \\ & = & \left(\dfrac{p^2-4}{3}\right)^3 + \left(\dfrac{2p^2+1}{3}\right)^3\end{array}$$ and it is clear that the two summands on the right hand side are integers, since if $p > 3$, $p^2 \equiv 1 \pmod 3$.
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