Let $d=a^{1999}+b^{1999}+c^{1999}$, we are going to show that $d$ is a multiple of $6$ which means $\mid d\mid$ is a composite.

Firstly, $d$ is a multiple of $2$ because $$d \equiv a^{1999}+b^{1999}+c^{1999} \equiv a + b + c \equiv 0 \pmod{2}$$

Next, by Fermat's little theorem, we have $x^3\equiv x\pmod{3}$. This conclusion can also be reasoned by factorizing $x^3- x= (x-1)x(x+1)$ which is a product of three consecutive integers. One of them must be a multiple of $3$. Hence $x^3-x\equiv 0\pmod{3}$. It follows that $$\begin{array}{rl} d &\equiv a \cdot a^{1998}+b\cdot b^{1998}+c\cdot c^{1998}\\ & \equiv a \cdot a^{666}+b\cdot b^{666}+c\cdot c^{666} \\ &\equiv a \cdot a^{222}+b\cdot b^{222}+c\cdot c^{222} \\ &\equiv a\cdot a^{74} + b\cdot b^{74} +c\cdot c^{74} \\ &\equiv a^{75}+ b^{75}+ c^{75} \\ &\equiv a + b + c \\ &\equiv 0 \pmod{3} \end{array}$$

This means $d$ is a multiple of $3$. Therefore, it must be a multiple of $2\times 3=6$.