Let the grid be
abc
dSe
fgh
with a,b,c,d,e,f,g,h being consecutive integers not neccisarily in order.
now we have
S=2(a+c+f+h)+b+d+e+g (1)
but if we rotate then we have
bce
aSh
dfg
and then we have
S=2(b+e+g+d)+(c+a+h+f) (2)
taking (1)(2) we end up with
a+c+f+h=b+d+e+g (3)
now let the consecutive integers be
x,x+1,x+2,x+3,x+4,x+5,x+6,x+7
by taking
a=x c=x+7 f=x+1 h=x+6 and
b=x+2 d=x+5 e=x+3 g=x+4 then we have both sides of (3)
equal to 4x+14
In general, all you need for this to work is to have the sum of the corner numbers be equal to the sum of the orthogonal numbers.

Posted by Daniel
on 20090830 14:05:42 