Question: Suppose abc could satisfy the given condition. Since a^2-1 is divisible by b, the numbers a and b are relatively prime. Thus, c^2-1 that is

Suppose abc could satisfy the given condition. Since a^2-1 is divisible by b, the numbers a and b are relatively prime. Thus, c^2-1 that is divisible by a and b, must be the common multiple of ab, which c^2-1ab. Since ac and bc, it follows that (abc^2). Therefore, it proves that such integers do not exist that are divisible by each of the others

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