Hellenic Mathematical Olympiad

Let $2007 + 4v = \kappa^2$, $\kappa \in \mathbb{N}$. Then
$$\kappa^2 - 4v = 2007 \Leftrightarrow \kappa^2 - 2^{2v} = 2007 \Leftrightarrow (\kappa - 2^v)(\kappa + 2^v) = 1 \cdot 3 \cdot 3 \cdot 223.$$
Since $\kappa - 2^v < \kappa + 2^v$, last equality is equivalent to the systems
$$\begin{cases} \kappa - 2^v = 1 \\ \kappa + 2^v = 2007 \end{cases} \text{ or } \begin{cases} \kappa - 2^v = 3 \\ \kappa + 2^v = 669 \end{cases} \text{ or } \begin{cases} \kappa - 2^v = 9 \\ \kappa + 2^v = 223 \end{cases} \quad (\Sigma_1) \text{ or } (\Sigma_2) \text{ or } (\Sigma_3)$$
Now we have
$$(\Sigma_1) \Leftrightarrow \begin{cases} \kappa - 2^v = 1 \\ \kappa + 2^v = 2007 \end{cases} \Leftrightarrow \begin{cases} 2\kappa = 2008 \\ 2 \cdot 2^v = 2006 \end{cases} \Rightarrow 2^v = 1003, \text{ absurd..}$$
$$(\Sigma_2) \Leftrightarrow \begin{cases} \kappa - 2^v = 3 \\ \kappa + 2^v = 669 \end{cases} \Leftrightarrow \begin{cases} 2\kappa = 672 \\ 2 \cdot 2^v = 666 \end{cases} \Rightarrow 2^v = 333, \text{ absurd.}$$
$$(\Sigma_3) \Leftrightarrow \begin{cases} \kappa - 2^v = 9 \\ \kappa + 2^v = 223 \end{cases} \Leftrightarrow \begin{cases} 2\kappa = 232 \\ 2 \cdot 2^v = 214 \end{cases} \Rightarrow 2^v = 107, \text{ absurd.}$$
Therefore, there are not natural numbers $v$ such that $2007 + 4v$ is a perfect square.