Problem 1. In the plane, three distinct fleas with integer coordinates are given. At every tick of time, two fleas sitting at the points $P$ and $Q$ can jump to two vacant points $R$ and $S$, such that $PQRS$ is a square (in that order of vertices). If three initial coordinates of fleas are $(0, 0)$, $(2, 0)$ and $(4, 1)$, prove that fleas can never become collinear.
More generally, find the set of all possible values missed by $2S$, where $S$ is the area of a triangle spanned by three fleas.
I intended to proposed the first part of this problem to IMO, but it turns to be too complicated.
Comments. Let $\alpha, \beta, \gamma$ be side lengths of a lattice triangle, $a=(\beta^2+\gamma^2-\alpha^2)/2$, $b=(\gamma^2+\alpha^2-\beta^2)/2$, $c=(\alpha^2+\beta^2-\gamma^2)/2$, and $s=2S$ (we also keep track of the sign of $s$).
The relation between these integers is given by $s^2=ab+ac+bc$. The jump of two fleas translates into the language of quadruples $(s;a,b,c)$ as one of six transformations
$(s;a,b,c)\rightarrow (s+b+c;a+b+c+2s,b,c)$,
$(s;a,b,c)\rightarrow (s-b-c;a+b+c-2s,b,c)$,
and 4 cyclic ones. We start from the collection $(2; 8, 9, -4)$. With MAPLE I was able to calculate that, for example, only these values of $|s|\leq 20000$ can possibly be missed:
$0, 1, 4, 5, 9, 11, 16, 25, 29, 35, 41, 55, 60, 64, 65, 80, 89, 101, 109, 125, 171, 191, 229, 275, 295, 299, 301, 315, 316, 336, 365, 380, 405, 476, 589, 595, 624, 635, 671, 689, 1055, 1109, 1225, 1495, 1955, 2171, 2215, 2315, 2699, 2761, 5225, 7471, 7555, 11365, 14845.$
Yet, I was not able to advance any further. At the moment, I am still not able to rigorously demonstrate that the given three points can never become collinear. In other words, that $0$ is indeed a member of the above sequence.
Addition 1. The sequence written above is obtained by the following procedure. Computer calculates $g$ generations of these $6$ transformations. If we work with arrays, the total number of configurations in the $g$-th generation would be $6^g$. In MAPLE though, it is much faster to work with sets due to considerable overlap, as well as the fact some transformations take backwards. Thus, $10$ generations were calculated (it took several hours, but optimization is still possible in order to push this to $12$ or $13$ generations).
Second, consider, for example, an ordinary member $(s; a, b, c)=(-37; 9, 20, 41)$ of the third generation. The transformations show that not only $s=-37$ is attainable value for $s$, but also all $s$ such that $s\equiv -37\equiv 21\text{ (mod }29)$, $s\equiv-37\equiv 13\text{ (mod }50)$, or $s\equiv -37\equiv 24\text{ (mod }61)$. We thus can filter out not only terms of the given generation, but all the corresponding arithmetic progressions.
And third, terms of the above sequence seem to stabilize rather quickly. For example, for $g\geq 7$ (it takes only several seconds to compute) all terms of the above sequence with terms $\leq 1000$ stabilize. It gives a true belief that $s=0$ is not possible, and that the majority of the numbers written are truly not possible values for $s$.
Addition 2 (a small progress). We may start from an even more compact triple of points, for which the answer is not trivial. In particular, consider $(0,0)$, $(2,1)$ and $(3,2)$ (the diameter is $\sqrt{13}$, as compared to $\sqrt{17}$ for the initial example). This corresponds to the quadruple $(s; a,b,c)=$ $(1; -3,5,8)$. We calculate also $10$ generations. All missed $s$ are even, and the list of missed $S=s/2\leq 1000$ is as follows:
$0, 1, 4, 5, 9, 16, 25, 29, 36, 49, 64, 80, 81, 99, 100, 121, 144, 169, 179, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961.$
But these are exactly integer squares, apart from five sporadic terms $5, 29, 80, 99, 179$.
Problem 2. Prove that, if we start from the points $(0,0)$, $(2,1)$ and $(3,2)$, then three fleas never span a triangle of area $N^2$, $N\in\mathbb{N}\cup\{0\}$.
It is not clear will sporadic values will eventually filter out, or these are truly sporadic ones.
