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Question Number 70997 by A8;15: last updated on 10/Oct/19

Answered by MJS last updated on 10/Oct/19

$$\mathrm{trying}\mathrm{first}$$$$\mathrm{for}\mathrm{an}\mathrm{easy}\mathrm{solution}\mathrm{both}a+24\mathrm{and}a+12$$$$\mathrm{must}\mathrm{be}\mathrm{square}\mathrm{numbers}$$$$\Rightarrow a=-8\Rightarrow x=-2$$

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