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The image contains $16$ starting values. The remaining cells can be calculated by arithmetic operations, keeping in mind that the line sum is $65$. Now is there an arrangement where I need less starting values to full resolve the grid. magic

Peter Woolfitt
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MANuNITED
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  • x x x x o / x x x x o / x x x x o / x x x x o / o o o o o This sqaure could be resolved if all x varialbes were defined. Now is there an arrangments were i need less than 16 starting values – MANuNITED Sep 07 '15 at 16:29
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    I do not know the minimum number of entries necessary to have a unique solutions, but I am quite sure, that $16$ is not the minimum number. – Peter Sep 07 '15 at 16:34
  • I am wondering if the middle cell is allways defined as 13, in the same way that a 3 by 3 sqaure must contain 5 in the middle cell – MANuNITED Sep 07 '15 at 16:37
  • No, the middle entry can be any number. – Peter Sep 07 '15 at 16:38
  • You can even arrange numbers in a large square, such that any $5x5$-square in it is a magic square. For this, you need so-called pan-magic squares (Look here : http://mathworld.wolfram.com/PanmagicSquare.html) – Peter Sep 07 '15 at 16:39
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    • x + x + x x x x x x Seems to be the minimum.

    – MANuNITED Sep 07 '15 at 16:42
  • I do not know, if the solution is unique in this case. – Peter Sep 07 '15 at 16:44
  • The least field hypothesis implies that least field for order n= n^2 - 2n+1. How would one prove this – MANuNITED Sep 07 '15 at 16:45
  • Probably only with exhaustive search. It was quite difficult to solve a related problem : How many entries are necessary to construct a sudoku with a unique solution. The answer was $17$, and the method to get this was simply brute force. – Peter Sep 07 '15 at 16:47
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    A naive analysis: We have $12$ equations and $26$ variables, so we could expect some way to do it by defining $14$ starting values. – Peter Woolfitt Sep 07 '15 at 16:48
  • @Peter consider the rank of the matrix. But that should be a good start. – Peter Sep 07 '15 at 16:50
  • @manu is it really conjectured that the number of clues necessary to produce a $n\ x\ n$-square which can uniquely be filled up, is $(n-1)^2$ ? – Peter Sep 07 '15 at 16:53
  • this seems to be the case for n= 1 2 3 – MANuNITED Sep 07 '15 at 17:02
  • assuming i know the centre value howmany leastfield must there exist – MANuNITED Sep 07 '15 at 17:06