Maclaurin 2018 Q1

Let the numbers be \(a\) and \(b\). Then \(a^2 + b^2 = 15(a + b)\) and \(a^2 - b^2 = \pm 3(a - b)\).
We do not know whether \(a\) and \(b\) are positive or negative, so if, for instance, \(a = 4\) and \(b = 3\), the expressions \(a^2 - b^2\) and \(a - b\) are both positive, but if \(a = -4\) and \(b = 3\) the first is positive and the second is negative. The ± sign is to take care of all alternatives. In fact, we have \(a^2 + b^2 \geqslant 0\) in the first equation, so we know that \(a + b \geqslant 0\). Now, by difference of two squares, we can write \(a^2 - b^2\) as \((a - b)\,(a + b)\) in the second equation. Hence, if \(a - b \ne 0\), we can cancel this term to obtain \(a + b = \pm 3\), and so we know that the positive sign is appropriate.
We now have \(a + b = 3\) and \(a^2 + b^2 = 45\). Writing \(b = 3 - a\), we obtain the quadratic \(a^2 - 3a - 18 = 0\) and solutions \((6, -3)\) and \((-3, 6)\), so the numbers are 6 and \(-3\) in either order.
However, there is also the possibility that \(a - b = 0\), but if that is the case the first equation becomes \(2a^2 = 30a\) and so \(a = 0\) or 15. Hence the pair of numbers is either 0 and 0, 15 and 15 or 6 and \(-3\).