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We have a square matrix, that all elements on main diagonal are zero, and other elements are following:

$$a_{i,j}=\begin{cases} 1,&\text{if i+j belongs to Fibonacci numbers,}\\ 0,&\text{if i+j doesn't belong to Fibonacci numbers}.\\ \end{cases}$$

We know that when $n$ is odd the determinant of this matrix is zero.

Now prove that when $n$ is even the determinant of this matrix is $0$ or $1$ or $-1$. (Use induction or other methods).

Also posted on MO.

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how can i prove that ? –  Zoha Shams Feb 11 '13 at 17:22
Hint: write out the 2x2 and 3x3 cases. Prove your statements for those. Next, use induction to prove the general case. Regards. –  Amzoti Feb 11 '13 at 17:24
i know that i should use induction, i ask you how can use induction?could u prove it? –  Zoha Shams Feb 11 '13 at 17:26
Posted a few days ago to MO, mathoverflow.net/questions/121243/… --- best to check the progress there before spending too much time on it. –  Gerry Myerson Feb 12 '13 at 5:34
@julien: Took a quick look and these are the things I have so far: matrix is always symmetric, rank is always even, n-odd $\det = 0$, sum eigenvalues = trace matrix = $0$, product of the eigenvalues = $\det = 0, 1, -1$ (for n-even, it is magnitude 1). Have not made more progress. You? –  Amzoti Jul 13 '13 at 20:29
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