# Magic square solving

I'm trying to solve a 3x3 magic square for 99 which starts at 29. I got the rows, columns and one diagonal but the other diagonal is (way) off.

Is there even a magic square which satisfies both diagonals for these parameters? As you can see I've been playing around in a spreadsheet, moving numbers around. But is there a quicker way to spot if a magic square for a number starting at an arbitrary number is possible?

• For a $3\times 3$ magic square to be filled with consecutive integers, there's not much choice. Let the numbers be $-4,-3,\dotsc,3,4$ and try to find which pattern is the only possible (except for rotations and reflections). – Daniel Fischer Oct 26 '14 at 22:07
• Hmm, i just solved it using another method which gives me the center number... – Ropstah Oct 26 '14 at 22:08
• Adding all number in the square and dividing it by 3 and again by 3 give me 33. Rest was a little puzzling. Thanks. – Ropstah Oct 26 '14 at 22:11

One magic square that works is

$$\begin{matrix} 32&31&36\\37&33&29\\30&35&34 \end{matrix}$$

It might interest you that one of Ramanujan's first hobbies was trying to understand magic squares, and they appear in his first notebook often.

Construction

I always construct it in following manner and it works, I don't know whether there is alternative way to do this or what's proof of correctness of my construction.

$a_1$ is starting number

$$\begin{matrix} a_8 & a_1 & a_6 \\ a_3 & a_5 & a_7 \\ a_4 & a_9 & a_2 \\ \end{matrix}$$

• For 3×3 squares, this is the only possible arrangement, not counting rotated and reflected versions of the same. – MJD Feb 28 '15 at 13:46

This is an odd*odd magic square. The arrangement would be as such:-

36 29 34
31 33 35
32 37 30

Start at the centre of the first row with the smallest digit. Go one row above and shift to one position right. If the position is already occupied place the digit below the first digit.(35 is placed below 34 as 32 occupies the required position.Similarly 32 is placed below 31 for the same reason.)