Proof for Ramanujan's Oblong magic square

Question

In his first notebook, Ramanujan discusses a $3 X 4$ magic square which he calls oblongs. In this he suggests that following would be the elements of a magic square: $$ \begin{array} {|r|r|r|} \hline A& C+D&A+2D&C+3D \\ \hline B+6D& B+4D& B+2D& B \\ \hline C& A+D& C+2D& A+3D \\ \hline \end{array}$$

where $A,B,C,D$ are all positive integers.I am not able to get the proof behind this solution.

Kindly give suggestions as to how I can prove this.

Where is the square? What is the magic? – Moti Aug 23 '19 at 05:11 — Moti, Aug 23 '19 at 05:11
Are $A, B, C, D$ all integers, positive integers or what? – Toby Mak Aug 23 '19 at 06:56 — Toby Mak, Aug 23 '19 at 06:56
yes I just edited my question – Deepa Kapoor Aug 23 '19 at 08:31 — Deepa Kapoor, Aug 23 '19 at 08:31

score 1 · Answer 1 · answered May 24 '22 at 07:59

1

Just to show what we are talking about (which has -for whatever reason - not been shown in the OP). Here a screenshot from BC Berndt's collection of Ramanujan's first notebook (pg. 20):

(Bruce C. Berndt: Ramanujan's Notebooks, Part I, Springer, 1985)

answered May 24 '22 at 07:59

Gottfried Helms

34,920

score 0 · Answer 2 · answered May 23 '22 at 00:45

You can't prove it because it isn't true for general whole number values of $A,B,C,D$. The columns all properly sum to $A+B+C+6D$, but two of the rows sum to $2A+C+6D$ and the other row sums to $4B+12D$. Presumably Ramanujan was saying there exist some integers that render the row sums equal and the oblong fully magic.

By the way, we cannot ever form a magic oblong with consecutive whole numbers unless the length and width of the rectangle are either both even or both odd.

Proof for Ramanujan's Oblong magic square

2 Answers2