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In this recent unanswered question I had asked about the probability of exactly one negative solution in a $3\times 3$ system of equations with Fibonacci-like coefficients, denoted by $F_i$.

With $\theta_i$ as defined in that question; that is, the concatenation of $1.F_i$, how can we prove that the sign of $F=\theta_i\theta_{i+5}-\theta_{i+1}\theta_{i+4}$ is positive only when $5\mid i$?

If this is disproved, is there an alternative pattern that $\text{sgn}(F)$ satisfies?

Note that I have defined $F_2=2$ not $1$ for the purpose of having different values for each $i$.

I ask this because when we find the inverse of that $3\times 3$ matrix, we end up with the expression in the denominator for $X,Y,Z$ so if we want to explore the case $X,Y\ge0$ and $Z<0$, we need to know its sign to progress on with removing the denominator.

Here are some numerical details I've found: \begin{array}{c|c}i&1&2&3&4&5&6&7&8&9&10\\\hline\operatorname{sgn}(F)&-&-&-&-&+&-&-&-&-&+\end{array}

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  • $\begingroup$ my guess: it is negative only when $F_{i+1}$ has more digit than $F_{i}$, that happens for $i=5,10,15$, for example. You should test it for $i=19$, since it is the first time that happens for non-multiple of 5 index $\endgroup$ – Exodd Sep 16 '18 at 10:46
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For $i=19$ where $5\not\mid 19$, $F$ is positive.

If this is disproved, is there an alternative pattern that $\text{sgn}(F)$ satisfies?

I have not seen any pattern for $i\le 30$.

\begin{array}{c|c}i&11&12&13&14&15&16&17&18&19&20\\\hline\operatorname{sgn}(F)&-&-&-&-&+&-&-&-&+&-\end{array}

\begin{array}{c|c}i&21&22&23&24&25&26&27&28&29&30\\\hline\operatorname{sgn}(F)&-&-&+&+&-&-&-&-&+&-\end{array}

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