Enigma 1050: Find the link
6 August 2018
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From New Scientist #2206, 2nd October 1999 [link]
I was trying to construct a chain of 2-digit and 3-digit perfect squares such that each square in the chain had at least two digits in common with each of its neighbours (or with its sole neighbour, if it was at either end of the chain). If a square had a repeated digit that digit only counted more than once in calculating the number of digits that the square had in common with another square if it also appeared more than once in the other square; so 121 had only one digit in common with 100, but 100 had two [digits] in common with 400.
I found that I could construct three totally different chains each consisting of at least five squares; and I made each of these chains as long as possible, consistent with the stipulation that no square should be used more than once. But I could not link these chains into a single long chain until I used one particular 4-digit square as the means of linking one end of my first chain to one end of my second chain and also the other end of my second chain to one end of my third chain. In each place where it was used, this 4-digit square had at least two digits in common with each of its chain neighbours.
(1) Identify this 4-digit square.
(2) Which were the squares at opposite ends of the single long chain?