The ordered pair of four-digit numbers (2025, 3136) has the property that each number in the pair is a perfect square and each digit of the second number is 1 more than the corresponding digit of the first number. Find, with proof, all ordered pairs of five-digit numbers and ordered pairs of six-digit numbers with the same property: each number in the pair is a perfect square and each digit of the second number is 1 more than the corresponding digit of the first number.

This is the second question from this year’s USAMTS, round one. Contest is over now, so post and discuss all you want.

It’s been like a month since I’ve last posted a math problem, so take some time to ponder upon this problem. It’s actually not that hard, and there’s a simple, systematic way of finding the solutions.

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