$ x^3(\mod 7) = \{ - 1,1,0\}$ we prove that $ n > 2$ then $ a_n$ can't be a square. We has $ 7\not |a_n$ Consider the equation: $ x^2 = y^3 + 1999$ $ x^2(\mod 7)\in \{0,1,4,2\}$ $ y^3 + 1999 (\mod 7)\in \{3,5\}$ So it has no solution when $ n > 2$ Now we prove that at most one of $ (a_1,a_2)$ can be a perfect square. Case 1$ a_1$ is a perfect square $ a_1 = m^2$ Suppose $ a_2$ is a perfect square then $ m^6 + 1999 = x^2$ (contradiction!) So if $ x_1$ is a perfect square then $ x_2$ is not a perfect square . So at most one of term in this sequence can be a perfect square.

doesn't '$ m^{6} + 1999 = x^{2}$ (contradiction!)' need a little more explanation? Anyway, my solution using a different modulus: If $ {a_{n}} = 1 (mod 4),$ then $ {a_{n + 1}} = 0 (mod 4)$ If $ {a_{n}} = 0 (mod 4),$ then $ {a_{n + 1}} = 3 (mod 4)$ If $ {a_{n}} = 3 (mod 4),$ then $ {a_{n + 1}} = 2 (mod 4)$ If $ {a_{n}} = 2 (mod 4),$ then $ {a_{n + 1}} = 3 (mod 4)$ Since squares are never 2,3 (mod 4), we must have that both $ {a_{1}}$ and $ {a_{2}}$ are squares to contradict our theorem. Assume $ {a_{1}} = x^{2}$ and $ {a_{2}} = y^{2}.$ Then: $ (y + x^{3})*(y - x^{3}) = y^{2} - (x^{2})^{3} = 1999$ . Since 1999 is prime, $ y - x^{3}$ must be 1. So: $ 1999 = (y + x^{3})*(y - x^{3}) = 2*x^{3} + 1$ $ 999 = x^{3}$ ; impossible for an integervalued x