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Cube + Single Digit = Perfect Square (Posted on 2024-01-21)

Which is the highest cube n^3 < 10⁸ that can be incremented by a single digit from 1 to 9 to yield a perfect square?

For example, 8 is a perfect cube. When incremented by 1 it becomes 81, which is a perfect square.

See The Solution Submitted by K Sengupta

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