|Ramanujan's genius (r) was discovered by Hardy (l)|
C+Q | A+P | B+R
A+R | B+Q | C+P
B+P | C+R | A+Q
where A,B,C are integers in arithmetic progression and so are P,Q,R.
Rewriting Ramanujan's scheme somewhat to
where P,Q,R are in the Rationals, it is clear that every (P,Q,R) yields a magic square with constant number 3 (P + Q + R).
I conjecture that for any 3 by 3 magic square a triple (P,Q,R) can be found in the Rationals such that they fit the above scheme. Finding a proof for this is one of my 'problems'. Naturally, I would be very interested in any counter-example.