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 $r=27$
$$11,1,15$$
$$13,9,5$$
$$3,17,7$$
Best Answer
Start by considering the constraints of the magic square. I denote the sum by $s$ rather than $a$ to avoid confusion. The sum of diagonals is $$ a_{11}+a_{13}+a_{31}+a_{33}+2 a_{22}=2 s$$ The sum of the first and last row minus the second column is $$ (a_{11}+a_{12}+a_{13})+(a_{31}+a_{32}+a_{33})-(a_{12}+a_{22}+a_{32})=a_{11}+a_{13}+a_{31}+a_{33}-a_{22} =s$$ From these equations you get $a_{22}= s/3$, in your case $a_{22}=1$.
Fix now two entries that are not on the same diagonal, let say $a_{11}$ and $a_{13}$. Given the constraints, all other entries are now fixed.
EDIT
We have $a_{22}=s/3$ and we fix $a_{11}=b$, $a_{13}=b'$ ($b$ and $b'$ are names that i give to recall that these entries are fixed).
The entry $a_{33}$ is fixed due to the constraint on the diagonal $$a_{11}+ a_{22}+a_{33}= s\Rightarrow a_{33} = s-\frac s3-b=\frac{2s}{3}-b$$ Similarly, $a_{31}=\frac{2s}{3}-b'$
We now have all the corner entries. This fixes all the other entries. For example, $$a_{12}=s-a_{11}-a_{13}=s-b-b'$$ Similarly, $$a_{21}=\frac{s}{3}-b+b', \quad a_{23}=\frac{s}{3}+b-b',\quad a_{32}=b+b'-\frac{s}{3}$$
(I let you verify that all the constraints are indeed verified.)