Austrian Mathematical Olympiad

$$\begin{array}{rl}{a}^2+a& =b^2+bc\\ 4a^2+4a+1& =4b^2+4bc+1\\ (2a+1{)}^2& =4b^2+4bc+1.\end{array}$$ Assume to the contrary that $c<1$ holds. This yields $$(2b{)}^2=4b^2<(2a+1{)}^2=4b^2+4bc+1<4b^2+4b+1=(2b+1{)}^2.$$ This is a contradiction as the square of an integer cannot lie strictly between two consecutive square numbers. Therefore, $c\ge 1$ holds (for instance, $a=b$ yields $c=1$ and therefore there is a solution of the equation with $c\ge 1$).