24th Hellenic Mathematical Olympiad

Let $2007+4v={\kappa }^2$, $\kappa \in \mathbb{N}$. Then $${\kappa }^2-4v=2007\iff {\kappa }^2-2^{2v}=2007\iff (\kappa -2^v)(\kappa +2^v)=1\cdot 3\cdot 3\cdot 223.$$ Since $\kappa -2^v<\kappa +2^v$, last equality is equivalent to the systems $$\{\begin{array}{l}\kappa -2^v=1\\ \kappa +2^v=2007\end{array}\text{ or }\{\begin{array}{l}\kappa -2^v=3\\ \kappa +2^v=669\end{array}\text{ or }\{\begin{array}{l}\kappa -2^v=9\\ \kappa +2^v=223\end{array}({\Sigma }_1)\text{ or }({\Sigma }_2)\text{ or }({\Sigma }_3)$$ Now we have $$({\Sigma }_1)\iff \{\begin{array}{l}\kappa -2^v=1\\ \kappa +2^v=2007\end{array}\iff \{\begin{array}{l}2\kappa =2008\\ 2\cdot 2^v=2006\end{array}\Rightarrow 2^v=1003,\text{ absurd..}$$ $$({\Sigma }_2)\iff \{\begin{array}{l}\kappa -2^v=3\\ \kappa +2^v=669\end{array}\iff \{\begin{array}{l}2\kappa =672\\ 2\cdot 2^v=666\end{array}\Rightarrow 2^v=333,\text{ absurd.}$$ $$({\Sigma }_3)\iff \{\begin{array}{l}\kappa -2^v=9\\ \kappa +2^v=223\end{array}\iff \{\begin{array}{l}2\kappa =232\\ 2\cdot 2^v=214\end{array}\Rightarrow 2^v=107,\text{ absurd.}$$ Therefore, there are not natural numbers $v$ such that $2007+4v$ is a perfect square.