Thankful For Equations

Number Theory Level 3

22 12 18 87 88 17 9 25 10 24 89 16 19 86 23 11 \begin{array} { | l | l | l | l | } \hline 22 & 12 & 18 & 87 \\ \hline 88 & 17 & 9 & 25 \\ \hline 10 & 24 & 89 & 16 \\ \hline 19 & 86 & 23 & 11 \\ \hline \end{array}

Ramanujan's birthday in DD/MM/YYYY format is 22/12/1887. He found the Ramanujan Square, which is a magic square whose first row is the numbers 22, 12, 18, 87.

Given that my birthday is 16/6/1975, does there exist a Chung Square, which is a magic square whose first row is the numbers 16, 6, 19, 75?

A magic square is filled with distinct integers such that the sum of each row, column and diagonal are equal.

Bonus: Can you find a magic square for your own birthday?
(Note: People born in 01/01, or in the year 2020, would not have a magic square since these numbers are repeated. To counter this, we can relax the distinct integer condition in the first row)

No Yes

Chung Kevin by Chung Kevin , 45, Australia

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3 solutions

Any Chung Square has the form as:

16 6 19 75 x y z 116 − x − y − z t x − z + t − 59 175 − x − y − t y + z − t 100 − x − t 169 − x − y + z − t x + y − z + t − 78 x + t − 75 \begin{array} { | c | c | c | c | } \hline 16 & 6 & 19 & 75 \\ \hline x & y & z & 116-x-y-z \\ \hline t & x-z+t-59 & 175-x-y-t & y+z-t \\ \hline 100-x-t & 169-x-y+z-t & x+y-z+t-78 & x+t-75 \\ \hline \end{array}

For example,

16 6 19 75 50 26 29 11 45 7 54 10 5 77 14 20 \begin{array} { | c | c | c | c | } \hline 16 & 6 & 19 & 75 \\ \hline 50 & 26 & 29 & 11 \\ \hline 45 & 7 & 54 & 10 \\ \hline 5 & 77 & 14 & 20 \\ \hline \end{array}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

We have 12 unknowns and 8 (linearly independent) linear equations, which gives us the 4 degrees of freedom x , y , z , t x, y, z, t . So you've indeed described all possible Chung Squares!

Do you know if we can also force these 12 unknowns to be distinct integers? That seems reasonable to me, but I'm not certain.

Chung Kevin - 4 years, 4 months ago
Santiago Hincapie
Jan 31, 2017

Suppose my birthday is the x 1 / x 2 / x 3 x 4 x_1/x_2/x_3x_4 , now consider the following square x 1 x 2 x 3 x 4 x 4 x 3 x 2 x 1 x 2 x 1 x 4 x 3 x 3 x 4 x 1 x 2 \begin{array} { | l | l | l | l | } \hline x_1 & x_2 & x_3 & x_4 \\ \hline x_4 & x_3 & x_2 & x_1 \\ \hline x_2 & x_1 & x_4 & x_3 \\ \hline x_3 & x_4 & x_1 & x_2 \\ \hline \end{array}

As x 1 , x 2 , x 3 , x 4 x_1,x_2,x_3,x_4 are any numbers then there is a Chung Square and of course also exist a Santiago Square.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I realize that the condition of "not necessarily distinct" isn't great.

I'm thinking of changing the problem to just distinct integers, and ignoring people who have birthdays that repeat. Do you know of a Chung Square solution?

Chung Kevin - 4 years, 4 months ago
. .
Feb 14, 2021

The answer is Y e s \boxed{Yes} because it always exists whose first row is someone's birthday.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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