In this problem, I gave a partially completed magic square and had people figure out what the rest of the information is.

That led me to think about the minimum amount of information that we need to provide, in order to determine the magic square.

I approach the problem in the following way:

- There are 10 unknowns - 9 values of the squares, 1 of the magic sum
- We have 8 equations - 3 horizontal, 3 vertical, 2 diagonal
- The system of equations isn't linear independent because the 3 horizontal sum up to the same as the 3 vertical. But I believe that dropping one of these equations is sufficient for linear independence, so we have 7 equations.
- If so, we then have a unique solution when there are 7 unknowns, which means that we need to provide information on 3 values
- However, I do not see a way to provide only 3 values, for the magic square to be uniquely determined.

Any idea on how we can proceed? Am I missing an equation, or is there a way to only use 3 values?

And how about the \( 4 \times 4 \) case?

## Comments

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TopNewestI feel that for any nxn magic square, you only need n clues to get unique solution. – Saya Suka · 1 week, 2 days ago

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– Chung Kevin · 6 days, 20 hours ago

The \( 4 \times 4 \) magic square will need a lot more than 4 clues.Log in to reply

@Zee Ell also mentioned that "Side sum is thrice of center square", which follows from summing up the 2 diagonals, the center row and the center column, and then subtracting the 3 rows.

This implies that we have 1 more equation in that system which is useless. Hence, it's possible for the magic square to just supply 3 numbers (like if we knew the entries in the left most column), which would then give us the center square and hence everything else.

Thanks! – Chung Kevin · 2 weeks, 4 days ago

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Sir, if negative values are allowed; then I "THINK" that if the values of the central square and any 2 adjacent corners is given we can solve for a unique magic square. I would like you to make a random magic square and omit all values except three; the central and any two adjacent corners; then I will try to solve and we may verify this. – Yatin Khanna · 3 weeks ago

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I posted this problem that was inspired by Zee Ell's comment. – Chung Kevin · 2 weeks, 4 days ago

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