M1. The sum of the squares of two real numbers is equal to fifteen times their sum. The difference of the squares of the same two numbers is equal to three times their difference.
Find all possible pairs of numbers that satisfy the above criteria.
Solution
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 $\pm$ 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 $a-b \neq 0$, 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 $a^{2}-3 a-18=0$ 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 $2 a^{2}=30 a$ and so $a=0$ or 15 . Hence the pair of numbers is either 0 and 0,15 and 15 or 6 and -3 .