Selection Examination for Juniors

Let $v(v+3)=w^2$, $w\in \mathbb{Z}$. We assume that: $v=3\kappa$, $\kappa \in {\mathbb{N}}^{\ast }$. Then we have: $$v(v+3)=3\kappa (3\kappa +3)=9\kappa (\kappa +1)=w^2\iff 3\cdot 3\kappa (\kappa +1)=w^2\Rightarrow 3|w^2$$ from which, since 3 is a prime, it follows that $3|w$. $$\text{Hence }\kappa (\kappa +1)={\left(\frac{w}{3}\right)}^2,\text{ with }\frac{w}{3}\in \mathbb{Z},\text{ and thus we have}$$ $${\kappa }^2<\kappa (\kappa +1)={\left(\frac{w}{3}\right)}^2<(\kappa +1{)}^2\Rightarrow \kappa <\left|\frac{w}{3}\right|<\kappa +1,\text{ a contradiction.}$$ Hence $v$ is not a multiple of 3.

Alternative solution: Let $v(v+3)=w^2$, $w\in \mathbb{Z}$. Then we have $$\begin{array}{rl}{v}^2+2v+1\le v^2+3v<v^2+4v+4& \Rightarrow (v+1{)}^2\le w^2<(v+2{)}^2,w\in \mathbb{Z}\\ & \Rightarrow w=v+1\text{ if }w=-(v+1).\end{array}$$ It follows that $$v(v+3)=(v+1{)}^2\iff 3v=2v+1\iff v=1\ne \pi o\lambda .3.$$