# 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?

-
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? – Babak 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. – Babak 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