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.
- To show that this is optimal, in the above image, if we're given any 3 of those 4 values, we can uniquely determine the magic square.

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

## Comments

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TopNewest@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 · 3 months, 1 week ago

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Can you explain your third dot point? – Gennady Notowidigdo · 1 week, 6 days ago

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Then we know that \( S = A + B + C\), \( S = D + E + F\), \( S = G + H + I \) from the horizontal rows.

If we are also told that \(S = A + D+ G \) and \( S = B + E + H \), then we can conclude that \( C + F + I = ( A + B + C ) + ( D + E + F ) + ( G + H + I ) - ( A + D+ G) - ( B + E + H ) = S + S + S - S - S = S \).

As such, this equation isn't "necessary", and we can drop it. – Chung Kevin · 1 week, 5 days ago

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I feel that for any nxn magic square, you only need n clues to get unique solution. – Saya Suka · 3 months ago

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– Chung Kevin · 2 months, 3 weeks ago

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

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 months, 1 week ago

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I posted this problem that was inspired by Zee Ell's comment. – Chung Kevin · 3 months, 1 week ago

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I think I found the general solution for such square ، the formul for all the possible numbers which are satisfy the conditions of the square , all the possible numbers up to the infinity which give you same summation in all directions

Notice that in all directions the summation will be 3M+9 which is the unique relation for the unique arrangement to have same summation in all direction Put it in Excel program to see every thing ، put M= - 3 and see how beautiful the symmetry of the numbers when you get zero summation in all directions .... – Mahmood Alsaad · 1 month, 1 week ago

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– Yatin Khanna · 1 month, 1 week ago

Not all magic squares will be like that though all values of M will generate a magic square..Log in to reply

Thank you ... – Mahmood Alsaad · 1 month, 1 week ago

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3, 15, 12,

19,10 ,1,

8, 5, 17, – Yatin Khanna · 1 month, 1 week ago

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– Mahmood Alsaad · 1 month, 1 week ago

you are right my friend ، this is approval that my theory is wrong ، hhhhhhhhhhhhhhh ، thank you ، I appreciate that ....Log in to reply

– Calvin Lin Staff · 1 month, 1 week ago

With the added assumption that the numbers must be consecutive integers, I believe there is only 1 solution up to rotation / reflection.Log in to reply

Thankfx..... – Mahmood Alsaad · 1 month, 1 week ago

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