Let us rewrite the above equation in terms of powers of 2's: $4^{995} + 4^{1500} + 4^{n} = 2^{1990} + 2^{3000} + 2^{2n} = 2^{1990}[1 + 2^{1010} + 2^{2n - 1990}]$ (i) Focusing on the bracketed factor in (i), we observe that: $2^{2010} + 2^{2n - 1990} + 1 = (2^{505})^{2} + 2 \cdot 2^{2n - 1991} + 1$ (ii) and (ii) will be a square when $2n - 1991 = 505$, or $\fbox{n = 1248}$