19. Let the prime $p \equiv 1(\bmod 6)$. Prove: There must be integers $u, v$ satisfying
$$4 p=u^{2}+27 v^{2}, \quad u \equiv 1(\bmod 3)$$
(Hint:
$$\begin{aligned} 4\left(a^{2}-a b+b^{2}\right) & =(a+b)^{2}+3(a-b)^{2} \\ & =(2 a-b)^{2}+3 b^{2} \\ & \left.=(2 b-a)^{2}+3 a^{2}\right) \end{aligned}$$