The 16th Japanese Mathematical Olympiad - The Final Round

Consider an arbitrary integer $k$. Take a complex number $\alpha$ that meets ${\alpha }^2=k$. (For example, let $\alpha =\sqrt{k}$ if $k$ is nonnegative, and $\alpha =i\sqrt{-k}$ if $k$ negative.) It is easy verify the following equalities: $$\begin{gathered} (n+\alpha )(n+1-\alpha )=((n(n+1)-k)+\alpha ), \\ (n-\alpha )(n+1+\alpha )=((n(n+1)-k)-\alpha ). \end{gathered}$$ By multiplying these two formulae, we obtain an identity, $$(n^2-k)((n+1{)}^2-k)=((n(n+1)-k{)}^2-k).$$ In consequence, for any $k$, we can pick up any $n$ and let $(a,b,c)=(n,n+1,n(n+1)-k)$ so that the required equality holds. There are infinitely many $n$ and hence $(a,b,c)$. Therefore, every integer $k$ meets the required condition.