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?

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## 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!

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Actually, there is only one answer.

Let \(m_{1}\) and \(m_{2}\) denote the sums of the middle row and middle column, respectively, of a 3 x 3 square array of numbers. Let \(c _{1}\) and \(c_{2}\) denote the sums of the main diagonal and secondary diagonal, respectively. Lastly, let S denote the sum of all nine elements of the square. Then if x denotes the centre element of the square,

\(m_{1}+m_{2}+c_{1}+c_{2} = S + 3x\)

Suppose the common sum is r,

x= r/3

Now \(a + r/3 + b = r\) , where a and b are elements in the square containing the middle element x,

Therefore, \(a + b = 2r/3 = 2x \)

=> a,x,b are in Arithmetic progression

That means the elements of diagonals and middle row and column are in AP.

Let the square be like this

7 | y | 12

w | x | t

8 | z | 13

\(19+y=21+z\) \(=> y=2+z\)

\(2x = y+z\) \(=>x= z+1\)

\(15+w = 25 +t\) \(=> w=10+t \)

\(2x = w+t\) \(=> x=5+t\) => \(z=4+t\)

\(x=(12+8)/2\) = \((7+13)/2 =10\)

Which gives us the square

7 | 11 | 12

15 | 10 | 5

8 | 9 | 13

with common sum equal to 30.There is no other answer, there may be more solutions but answer is one only.

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You had most of it. Just note that

I've edited the first few equations so that you can take a look at it.

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Thanks

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Some pointers to note

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Can you explain your third dot point?

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If the magic square was

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.

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I feel that for any nxn magic square, you only need n clues to get unique solution.

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

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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.

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Negative values are allowed, as are fractional / decimal values. It shouldn't affect the square, because we can just "add k" to all terms.

I posted this problem that was inspired by Zee Ell's comment.

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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 ....

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

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Yes ....... but under this condition to have the same summation in all directions I think there is No other arrangement than this ....

Thank you ...

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3, 15, 12,

19,10 ,1,

8, 5, 17,

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

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Actually it depends on the value of M ، just insert M and see the corresponding set of values ...if M is integer number the set of corresponding numbers will be integer also ....

Thankfx.....

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