Mock AIME I 2015 Problems/Problem 9
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Since is a multiple of , let . We can rewrite the first and second conditions as: (a) is a perfect square, or is a perfect square. (b) is a power of , so it follows that , , and are all powers of . Now we use casework on . Since is a power of , is or or . If , then no value of makes . If or , then no value of that is a power of makes a perfect square. If , then and for solutions. If , then and for solutions. If , then and for solutions. If , then and for solutions. This is a total of solutions.