Ramanujan's method for solving a Magic Square Ramanujan came up with a formula for solving a magic square.

Theorem 1: Let $m_1,m_2$ denote the sums of the middle row and middle column respectively for a $3\times 3$ square array of numbers. Let $c_1,c_2$ denote the sums of the main diagonal and secondary diagonal respectively. And lastly, let $S$ denote the sum of all nine elements of the square. If $x$ denotes the center element, then$$x=\frac 13\left(m_1+m_2+c_1+c_2-S\right)\tag1$$
Suppose that the sum of each row and column is equal to $r$. Then, we have$$x=\frac 13\left(c_1+c_2-r\right)\tag2$$
And note that that implies that $x=\frac r3$, so $r$ is a multiple of $3$.
And lastly, in a $3\times 3$ magic square, the elements in the middle row, middle column and diagonals are in arithmetic progression. More clearly, let $a,b$ be the first and third elements respectively. Then$$a+r/3+b=r\tag3$$Hence$$b-r/3=r/3-a\tag4\\\vdots$$


Although the formula is clear, I'm just wondering how you would go about actually solving a magic square.

Question: Using Theorem 1, how would you solve a magic square? (Example provided below)


For example:


*

*Construct a magic square with $r=15$ and all numbers are odd!



My Attempt:  Since $r=15$, we have $x=15/3=5$. Thus, the middle digit$^{[1]}$ is $5$. From $(3)$, we get a Diophantine equation, and solving we get $$(a,b)=(1,9)\\(a,b)=(3,7)\\(a,b)=(5,5)\tag5$$
After that, I'm not sure what to do. I have $3$ possible $a,b$ values and I'm not sure which two I should use. Any help would be accepted!
$\scriptsize{[1]:\text{Maybe..?}}$
 A: Why it's impossible:
There are actually 4 more arithmetic sequences that will always be present in a magic square.
Let the square:
$$a,b,c$$
$$d,e,f$$
$$g,h,i$$
So from $(3)$ and $e=\frac{r}{3}$ we get the equations:
$$a-e=e-i$$
$$b-e=e-h$$
$$c-e=e-g$$
$$d-e=e-f$$
$(6)$ From these we get:
$$2e=a+i=b+h=c+g=d+f$$
Next take:
$$r=a+b+c=a+d+g\implies b+c=d+g$$
Then from $(6)$, substitute in $g=b+h-c$ to get:
$$b+c=d+(b+h-c)$$
Rearrange:
$$d-c=c-h$$
The same "trick" can be done too for $(b,g,f),(b,i,d),(h,a,f)$ either using the same method or the symmetries of the magic square. 
$(7)$ We have:
$$d-i=i-b$$
$$b-g=g-f$$
$$f-a=a-h$$
$$h-c=c-d$$
Note the middle elements are the corners, which is helpful for remembering these.
Back to your question now. 
For $r=15$ we have 4 sequences across the center: $(1,9),(2,8),(3,7),(4,6)$
From these we can make 5 more arithmetic sequences: $(1,2,3),(2,4,6),(1,4,7),(3,6,9),(7,8,9)$ 
All 5 progressions have at least one even number so no such square is possible!
Finding the smallest odd magic square:
Since the square with $5$ in the center (and a total of $15$) is impossible. Let's try the square with $7$ in the center. 
The sequences about 7 are $(1,13),(3,11),(5,9)$
From these we can pull the sequences $(1,3,5),(1,5,9),(9,11,13),(5,9,13)$
The middle elements will become the centers and the left/right elements will become the middles of the edges of the square. This implies $5$ and $9$ will be duplicated. So no distinct square exists with center element $e=7$
However for $e=9$ we can take the pairs about $9$: $(1,17),(3,15),(5,13),(7,11)$
Now note for the equations in $(7)$, each equation's last variable is the first variable of the equation following it. That is to say, we must pull 4 sequences from the elements of the pairs above such that they form a sort of loop with their first and last elements. The right sequence can be found with a small bit of work:
$$\rightarrow(1,7,13)\rightarrow(13,15,17)\rightarrow(17,11,5)\rightarrow(5,3,1)\rightarrow$$
Now we construct the square, starting with $(1,7,13)$:
$$a,[1],c$$
$$[13],[9],f$$
$$g,h,[7]$$
Then $(13,15,17)$:
$$a,1,[15]$$
$$[13],9,f$$
$$g,[17],7$$
Then $(17,11,5)$:
$$[11],1,15$$
$$13,9,[5]$$
$$g,[17],7$$
Finally $(5,3,1)$:
$$11,[1],15$$
$$13,9,[5]$$
$$[3],17,7$$
And we're done. The smallest all odd magic square has $r=27$
$$11,1,15$$
$$13,9,5$$
$$3,17,7$$
A: 
You can always arrange (rotate, reflect, etc.) the numbers in a magic square so that the smallest is top-middle and the next is bottom-right. Then there are two possible arrays of ranks of the $9$ numbers.
\begin{array}{cc}
   \text{TYPE 1} & \text{TYPE 0} \\ 
      \begin{pmatrix}
         8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2
      \end{pmatrix} 
   &
      \begin{pmatrix}
         8 & 1 & 7 \\ 4 & 5 & 6 \\ 3 & 9 & 2
      \end{pmatrix}
\end{array}
The first is itself a magic square, the second is not. The nicest examples of a TYPE$1$ and a TYPE $0$ magic square are
$$
   M_1 = \begin{pmatrix}
         8 & 1 & 6 \\ 3 & 5 & 7 \\ 4 & 9 & 2
         \end{pmatrix}
  \qquad
   M_0 = \begin{pmatrix}
         8 & 0 & 7 \\ 4 & 5 & 6 \\ 3 & 10 & 2
         \end{pmatrix}
$$

The first $9$ odd numbers are
$$1,3,5,7,\color{red}9,11,13,15,17$$
with $9$ in the middle. So the smallest magic sum has to be $3\times 9 = 27$.
That means that $r=15$ is not possible.
The smallest TYPE$1$ magic square would be
$$2M_1 - 1 = \begin{pmatrix}
         15 & 1 & 11 \\ 5 & 9 & 13 \\ 7 & 17 & 3
         \end{pmatrix}$$
The smallest TYPE$0$ magic square would be
$$2M_0 + 1 = \begin{pmatrix}
                17 & 1 & 15 \\ 9 & 11 & 13 \\ 7 & 21 & 5
             \end{pmatrix}$$
