Question Number 70997 by A8;15: last updated on 10/Oct/19
Answered by MJS last updated on 10/Oct/19
$$\mathrm{trying}\:\mathrm{first} \\ $$$$\mathrm{for}\:\mathrm{an}\:\mathrm{easy}\:\mathrm{solution}\:\mathrm{both}\:{a}+\mathrm{24}\:\mathrm{and}\:{a}+\mathrm{12} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{square}\:\mathrm{numbers} \\ $$$$\Rightarrow\:{a}=−\mathrm{8}\:\Rightarrow\:{x}=−\mathrm{2} \\ $$