Why is there only one 3x3 magic square?

A magic square has every row, column, and diagonal sum to the same number. How many magic squares are there using the numbers 1 to 9?

This video shows you all the possibilities and proves there are no others, leading to the surprising result the 3x3 magic square is essentially unique, up to symmetries of a square.

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[RoderickEtheria]

Move all the numbers once clockwise around the 5, and then put the even numbers in the opposite corners. Reflect and rotate the square to get all possibilities. The total of all 9 numbers is 45. Divide this by 3 rows makes 15. The possible ways to create 15 are 1+5+9, 1+6+8, 2+4+9, 2+5+8, 2+6+7, 3+4+8, 3+5+7, and 4+5+6. All other combinations don't add up to 15. The central square must fit in 4 different combinations. This requires the number in the center to be 5. This occupies the combinations 1+5+9, 2+5+8, 3+5+7, and 4+5+6, leaving the combinations 1+6+8, 2+4+9, 2+6+7, and 3+4+8. All of the corner squares must fit two of these combinations, which makes 2, 4, 6, and 8 the corners. You can fill the square based on this fact, and you will reveal there is only one real answer for a 3x3 magic square, with variations based on rotation or reflection.

[samuelturner]

there is an easy way I learned years ago. and i actually applied it on 10*10 square, the trick is you have to start on cell which is the last cell on right of the 2nd row from bottom. and after that the next no to put in will be 1 down and 1 right

1 2 3
A
B
C

you are on the cell B3 in this cell you will start and write 1 in it. now you will go 1 down and one left so one down will be C3, then 1 right will be C1 (as there is no cell, you have to go to the left most cell.) now in C1 write 2, and follow this rule, when you come to a cell which is already occupied then you must select the cell which is on the top of it. Assume you come to cell B2 which is already occupied then you select the cell B1.

Follow this pattern and you can solve it easily.

Once you understand the formula you can use this to solve any magic square

[Nofxthepirate]

Another interesting note is that all the pairs you can make out of 7, 8, and 9 sum to 15 or more with only two numbers. This means that 7, 8, and 9 must never be in the same row, column, or diagonal as each other. This leaves only 3 possibilities for where 7 8 and 9 can go. Also, 6 and 9 cannot be in the same row, column, or diagonal because they also add up to 15 with just two numbers. This set of restrictions was the first observation I made when trying to solve the question on my own. Unfortunately it's not enough information to solve the whole puzzle and it basically left me in a position of trying to do sudoku to figure out where to place the other numbers by process of elimination. The missing link for me was counting the number of times each other digit shows up in the possible calculations.

[juanitacamacho]

When I was about 13, I saw this puzzle without explanation. I took me a whole day to figure it out. Now I know the logic and must teach it to my grandkids.

[jacobk]

This is actually a very interesting approach to this kind of puzzle. You are more or less tackling some of the fundamentals of group theory in without explicitly saying so.

[seemachauham]

This is given by my teacher as a holiday assignment,

[maxzriver]

You can also form 3 * 3 squares with trios of constant ratio numbers
Example: 2 3 4
6 7 8
10 11 12

11 2 8
4 7 10
6 12 3

[jumpman]

Here's how I did:

1+2+3+4+5+6+7+8+9 = 45, so the sum of the elements in each triplet has to be 45/3 = 15.

If the lowest number in a triplet is 1, then the second lowest number has to be at least 5, giving us the highest number as 9.
If we then increase the middle number by one, giving us 6, we have to decrease the highest number by one (in order to maintain the sum of 15), giving us 8.
Repeating this process, the middle number becomes equal to the highest number, which is of course not an option, so all-in-all there are two possible triplets containing the number 1.

In the same manner we can show that there are only eight unique triplets:
(1, 5, 9), (1, 6, 8), (2, 4, 9), (2, 5, 8), (2, 6, 7), (3, 4, 8), (3, 5, 7) and (4, 5, 6)

Seeing as there are 3 rows, 3 columns and 2 diagonals (3+3+2 = 8), all eight triplets must be used in order to complete the magic square.

Let's start with the middle square. This square is connected to the middle row, the middle column and the two diagonals (2+2+1 = 4), so any number that goes here has to be present in at least four of the eight triplets, and the only number that applies to is 5.
So the middle square can only contain the number 5.

Now, the number 1 can only be put in the upper-left square or the upper-middle square (all other possibilities are either rotations or mirrors of these two scenarios).

If we choose to put it in the corner square, this gives us that the lower-right square has to contain the number 9, and the upper row has to be either (1, 6, 8) or (1, 8, 6):
1 6 8 1 8 6
5 5
9 9

Both scenarios make the sum of the numbers in the rightmost column to exceed 15, which means that the number 1 can only go into the upper-middle square.
This gives us that the lower-middle square has to contain the number 9, and the upper row has to be either (6, 1, 8) or (8, 1, 6), but these two scenarios are mirror images of each other, which leaves us with but one possibility:
6 1 8
5
9

This gives us that the diagonals have to be (6, 5, 4) and (8, 5, 2):
6 1 8
5
2 9 4

Then, the leftmost column has to be (6, 7, 2) and the rightmost column has to be (8, 3, 4), thus completing our magic square:
6 1 8
7 5 3
2 9 4

This can be rotated up to three times. It can also be mirrored and then that mirror image can be rotated up to three times. This gives us a total of 4×2 = 8 possible solutions:

6 1 8 2 7 6 4 9 2 8 3 4
7 5 3 9 5 1 3 5 7 1 5 9
2 9 4 4 3 8 8 1 6 6 7 2

8 1 6 4 3 8 2 9 4 6 7 2
3 5 7 9 5 1 7 5 3 1 5 9
4 9 2 2 7 6 6 1 8 8 3 4

[Utkarsh_Singh]

There is another way to find the center no.
Assuming the sum of each row/col/diagonal to be S(where S=15)
then
4S= 45(sum of 1-9)+ 3*centre
(as shown in video we can pass 4 lines from center and counting the sum of numbers traced by lines equals sum of all numbers plus 3*centre no)
now on evaluating the above eqn you can find the centre no.

[cndbrn]

I watched a similar digital root video earlier, so I decided to follow that a bit.
All the odd #'s have 2 of each, The even #'s have 3 of each, and the (5) has 4.
So there's 4 odd #'s - 4 even #'s and the 1 (5)
So I took 234, 441 / 9 = 26, 049
reversed it and did the same
441, 234 / 9 = 49, 026
Now I took 49, 026 - 26, 049 = 22, 977 the digital root of this is 7+2, 7+2, 9 = 999

[shellypeng]

Great video!
I figured out a quicker way: if we want to fill the number 9 in, we have a total of 6 left for the other 2 squares that are in row/column with 9(bcuz 9+6 =15, its the sum)since there are only 2 ways to add up to 6(1&5, 2&4), we can garantee that 9 has to be at where the T shape intersects. Then according to the middle number is 5, we can fill out the other number in row with 9 - which is 1. Then we take the other pair of numbers that add up to 6 - whcih is 2&4, we can fill out the column that has 9 in there. Then we just subtract the numbers we fill out and can easily fill out the dquare.without finding all 8 variations that gives 15.
Sorry for my English.

[starpawsy]

If you look at higher odd ones, like 5X5 or 7X7 or greater, you can immediately see the genertal solution.

[rohanbanerjee_]

thanks, 🙂 it is very needed for my coding problem in 2022

[_adityakumar]

Really Nice.Everywhere I got trick and shortcuts.But here I got mathematical approach.Wonderful Sir.Very nicely explained.

[briank.]

I just found this and realized a math competition I entered in had this same problem but used 5 in the middle like always and x above that and x+1 in the bottom right corner and asked for the sum of all possible x values. I just skipped that one :p

[rileystarnes]

This is the best explanation of this magic square that I have been able to find on YouTube. Thank you for taking time out of your day to provide free knowledge to the public! Good work.

[beatricendege]

Hi you are a good teacher hope you have a blessed day thank you dear

[_adityakumar]

This is really expected from you only Sir.

[leoniecartlidge-gann]

Thank you so much! Home schooling my child and this video has been a godsend! Extremely helpful and so very easy to understand. Your other book resources look awesome too! So glad I "stumbled" across your channel whilst performing a mathematical search

[sallaklamhayyen]

Wow, Brilliant explanation thank you so much
please continue making this kind of Math puzzles or do a course of Discrete Math