A magic square is a square which allow non-negative integers entries in which all row sums and columns sums are equal. Let $H_3(r)$ denotes number of magic squares of size $3*3$ in which each row and column have sum equals $r$. Prove that $$H_3(r) =\binom{r+4}{4}+\binom{r+3}{4}+\binom{r+2}{4}$$ where $H_3(r)$ is the number of $3*3$ magic squares of line sum $r$.

I try to find out the relations between all the $$ \left[ \begin{array}\\ a&b&c\\ d&e&f\\ g&h&i\\ \end{array} \right] $$ variables. I thought if I manage to built a relation between all the variables so that the values of all variables depend upon any one or two independent variables. But I fail to built such relations among variables.

I asked this question to all my faculty teachers and school teacher, but no answer come up.

If the question is worth, please solve it.

  • $\begingroup$ @MorganRodgers Yes sir I tried it, but the equations formed are dependent on one another. $\endgroup$ – Parker May 11 at 19:19
  • $\begingroup$ Up to what I see, there are $5$ conditions. [1-4] $\max(a+b, d+e, a+d, b+e ) \le r$ for $c,f,g,h \ge 0$ and [5] $a+b+d+e \ge r$ for $i \ge 0$. $\endgroup$ – achille hui May 11 at 19:27
  • $\begingroup$ Usually, the diagonals are also of the same sum in magic squares - I suppose here that's not the case? $\endgroup$ – vrugtehagel May 11 at 19:58
  • $\begingroup$ @vrugtehagel If you force the diagonals to also sum to $r$ then the formula will not give the right count. $\endgroup$ – Morgan Rodgers May 11 at 20:08
  • $\begingroup$ Anyhow, this equation counting the "magic squares" (missing the usual condition that diagonals also sum to $r$) seems to be correct, but I don't see an easy insight for giving a combinatorial proof. You probably want to interpret each of the three binomial coefficients as "counting" the number of squares of a particular form. $\endgroup$ – Morgan Rodgers May 12 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.