# Proving a certain determinant $\left|\det A\right|$ is complete square

Consider the following matrix $$A_{ij}= \begin{cases} 1\quad\text{ if }\space (i+j)\space\text{ is prime,}\\ 0\quad\text{ otherwise.} \end{cases}$$ How can one prove that $\left|\det A\right|$ is a complete square?

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I assume we are talking here about an $n\times n$ matrix, with the indices satisfying $1\leq i,j \leq n$? –  Old John Aug 30 '12 at 13:27
@OldJohn: Is the induction on $n$ applicable here? –  B. S. Aug 30 '12 at 13:30
probably, but I am not sure yet! –  Old John Aug 30 '12 at 13:32
@OldJohn: It seems that this matrix looks symmetric. –  B. S. Aug 30 '12 at 13:41
For $6 \times 6$, the determinant is $-1$. See here.. Ah, but he's taking the modulus.. Anyway, some numerical data: For $n =4,5,6$, the answer is $0,1,-1$. –  Rijul Saini Aug 30 '12 at 14:09

It was a problem posed at an annual contest for undergraduates called IMC (International Mathematics Contest). The official site of the site http://www.imc-math.org . It was the fifth problem of the second day (year 2008), which means it is pretty hard to solve. Here is a reference of the official site http://www.imc-math.org.uk/imc2008/day2_solutions.pdf . Also there is a discussion here at artofproblemsolving forum http://www.artofproblemsolving.com/Forum/viewtopic.php?f=79&t=217339 .

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I faces this problem on a mathematical battle at Saint-Petersburg. –  Norbert Aug 30 '12 at 14:36
I think there are too many competitions for the problems to be all original.. –  clark Aug 30 '12 at 14:42