\section*{Problem 5}

Are there any solutions in positive integers to \(\mathrm{a}^{4}=\mathrm{b}^{3}+\mathrm{c}^{2}\) ?


\section*{Solution}

We have \(\mathrm{b}^{3}=\left(\mathrm{a}^{2}-\mathrm{c}\right)\left(\mathrm{a}^{2}+\mathrm{c}\right)\), so one possibility is that \(\mathrm{a} 2 \pm \mathrm{c}\) are both cubes. So we want two cubes whose sum is twice a square. Looking at the small cubes, we soon find \(8+64=2\). 36 giving \(6^{4}=28^{2}+8^{3}\). Multiplying through by \(\mathrm{k}^{12}\) gives an infinite family of solutions. Note that the question does not ask for all solutions.

