Magic Square Connections Study

Prior to BRILLIAthon's last problem, I had made a study about magic square connections.

Now this is for \(n \geq 2\) as \(n = 1\) yields no connections.


Imagine a 3×33 \times 3 magic square without numbers:

Now, in a magic square, you need to make sure all the rows, columns and (sometimes) diagonals add up to one number.


Most people would see 88 connections in this picture. But I see differently. I see 1212 connections:

Why 1212? Why not see as everybody else?

Well, from what normal people see, the amount of connections in a magic square forms this arithmetic sequence:

0,6,8,10,12,14,...0, 6, 8, 10, 12, 14, ...

There's no nnth term.

And there is additional connections.

But, from my perspective:

0,6,12,18,24,300, 6, 12, 18, 24, 30

There's a nnth term for this sequence: 6n66n - 6.

And there is no additional connections.

Also, when n10n \geq 10, it gets harder to draw a magic square with all the connections, so we need an algebraic approach.

What people normally see is 2n+42n + 4 (excluding the 00).

However, the ratio 6n62n+4\frac{6n - 6}{2n + 4} shows, the amount of magic square connections missing gets larger with each increase of 11 to nn.

Also, the amount of triangles increases at a geometric rate:

In my square above, you should see (forgive me for the messy lines - it was done on PowerPoint) 1616 triangles. (If you can't see 1616 triangles, don't worry - I first did this study on paper, so I know exactly how many are there.)

Continuing the sequence in both directions, we obtain:

0,4,16,36,64,100,...0, 4, 16, 36, 64, 100, ...

The nnth term is: (oddly, 2n22n^2 or (2n)2(2n)^2 don't match the exact sequence.) (n1(4(n1))(n - 1(4(n - 1)).

Note that this nnth term is from my perspective.

Now, imagine functions MS(A)M_S(A) and MS(P)M_S(P) (whereMSM_S denotes the magic square, (A)(A) denotes area and (P)(P) denotes perimeter):

MS(A)=n2M_S(A) = n^2

MS(P)=4nM_S(P) = 4n

The ratios are:

P:A=P : A = n4\frac{n}{4}

A:P=A : P = 4n\frac{4}{n}

Based on these facts:

P:A=(A:P)1P : A = (A : P)^{-1}

A:P=(P:A)1A : P = (P : A)^{-1}

Now, if we divide the amount of triangles by the amount of connections (C:NTC : N_T for short), we get: 4(n1)6\frac{4(n - 1)}{6}

Inverting it gives us the amount of connections divided by the amount of triangles: 64(n1)\frac{6}{4(n - 1)}


NT:C=(C:NT)1N_T : C = (C : N_T)^{-1}

C:NT=(NT:C)1C : N_T = (N_T : C)^{-1}

Now, showing a 1×11 \times 1 square and increasing nn by 11 gives us this:

1,3,5,7,9,...1, 3, 5, 7, 9, ...

The nnth term for this sequence is 2n12n - 1 - this nnth term denotes the function AD(x)A_D(x) (where ADA_D is the amount of dots.)

Also, since a square's rotational symmetry is 44, a magic square with all the connections will also have a rotational symmetry of 44.

Now all of this information is useful for solving questions like this:

Example 11:

You have a n×nn \times n square.

You have one clue:

n8=n - 8 =n9\frac{n}{9}

Find the true amount of connections, the amount of dots, the amount of triangles and add them together.


n8=n - 8 =n9\frac{n}{9}

9(n8)=n9(n - 8) = n

9n72=n9n - 72 = n

8n72=08n - 72 = 0

8n=728n = 72

n=9n = 9

Substitute n=9n = 9 into:

(n1(4(n1)))(n - 1(4(n - 1))) - 256256

6n66n - 6 - 4848

2n12n - 1 - 1717

256+48+17=321256 + 48 + 17 = \fbox{321}

Now, to the last BRILLIAthon problem:

How many connections are there in a 20×2020 \times 20 square:


(6×20)6=6×19=114(6 \times 20) - 6 = 6 \times 19 = \fbox{114}

I hope you enjoyed this study and learnt something from this!

Note by Yajat Shamji
1 year ago

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@Levi Walker, check out my study!

First ever in my profile!

Yajat Shamji - 1 year ago

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Really nice! It seems that with more connections they would also lose their "magic" property where summing along a connection will lead to the same constant.

Levi Walker - 1 year ago

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I know. There is only one way that the ''magic'' property would stay (investigated this as well):

Putting 00 in every square!

Yajat Shamji - 1 year ago

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I think you should be more clear on what "connections" means. I did not know what you meant in the problem, and I had to not do your problem because of that.

Elijah L - 1 year ago

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Well, I think you understand now, right?

Yajat Shamji - 1 year ago

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No, not really.

Elijah L - 1 year ago

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