TITLE: Rectangle with coordinates of all vertices Fibonacci numbers
QUESTION [6 upvotes]: Suppose the coordinates of all vertices of a given (non-degenerate) rectangle are Fibonacci numbers. Suppose that the rectangle is not such that one of its vertices is on the $x$-axis and another on the $y$-axis. Is it true that  either the sides of the rectangles are parallel to the axes, or make an angle of $45$ degrees with the axes.

REPLY [2 votes]: That's true, for the reason that the Fibonacci sequence grows quite fast.
Assume that there exists a rectangle in $F\times F$, where $F=\{F_2,F_3,F_4,\ldots\}$, whose sides are not parallel to the axes, neither make an angle of 45 degrees with them. Let $(F_a,F_b),(F_c,F_d),(F_e,F_f),(F_g,F_h)$ be the vertices.
We can assume $a<c<e-1$, $d<b$, $f>d$ and 
$$ m = \frac{F_c-F_a}{F_b-F_d} \in (0,1) $$
by symmetry. Clearly, $g\leq e-1$ must hold, so, by considering the $(F_a,F_b)-(F_g,F_h)$ side, we have:
$$F_h \leq F_b + m(F_{e-1}-F_a) = F_d-\frac{1}{m}(F_c-F_a)+m(F_{e-1}-F_a).$$
However, by considering the $(F_g,F_h)-(F_e,F_f)$ side, we have:
$$F_h \geq F_f + \frac{1}{m}(F_e-F_{e-1}) = F_d+m(F_e-F_c)+\frac{1}{m}F_{e-2}.$$
This two inequalities are not compatible, since:
$$m(F_e-F_c)\geq m F_{e-1} > m(F_{e-1}-F_a),\qquad \frac{1}{m}F_{e-2}\geq 0\geq -\frac{1}{m}(F_c-F_a),$$
so the only possibilities in order to have a rectangle in $F\times F$ is to take a rectangle with sides parallel to the axis or with $m=1$.
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