Proof 1) Each frog will always have integer coordinates. In addition, their coordinates will remain constant mod 2. Proof: If a with integer coordinates (a , b) jumps over frog with integer coordinates (x , y), then its new coordinates are (2x-a , 2y-b), which are integers and congruent to (a , b) mod 2. In the beginning, the frogs have all the possible pairs of coordinates from {0, 1}^2, while a square with side 2, with vertices with integer coordinates, will have all vertices equal mod 2, so the frogs can’t reach such a formation.

Proof 2, which is kinda similar to 1, but cooler and longer) Each three frogs always span a triangle with integer coordinates, and therefore its area, times two, is always an integer (this is well known; NOTE: from now on, “area” will denote 2*area). All four of these areas start as 1.

Lemma A: At any point, the sum of these four areas mod 2 is 0.
Let us denote the frogs A,B,C,D. Up to changing the frogs’ names, there are two cases:
(Case I) ABCD is a convex quadrilateral. In this case, the sum of the four areas == (ABC + CDA) – (BCD + DAB) = (ABCD) – (ABCD) = 0 (mod 2) [where the strings of letters denote areas].
(Case II) D is a point inside triangle ABC. In this case, the sum of the four areas == (ABC) – (ABD +BCD + CAD) = (ABC) – (ABC) = 0 (mod 2).
In both cases we arrived at the conclusion that the sum == 0 (mod 2), QED.

Lemma B: The residues of the areas mod 2 never change. Indeed, if A jumps over B, then the residues of ABC, DAB and BCD do not change, and neither does the residue of ABC +BCD +CDA + DAB, which is always 0; therefore the residue of CDA does not change as well. So, all of the residues are always 1. We can now conclude that we cannot reach a square with side 2, for that will mean the area of each triangle is 4, which is not 1 modulo 2.

Neither of the above proofs would tell us if we could make the frogs into a square with an odd side length. The following simple proof shows why we can’t make any square with side length x > 1; but unlike the first two proofs, it only wields results about squares.

Proof 3) The map of the frogs movement is completely reversible. That mean that if we can get from a square with side 1 to a square with side x, we can get from a square with side x to a square with side 1. But, by applying the same jumps to an only trivially different starting formation, that would mean that we can get from a square with side 1 to a square with side 1/x, but it is obvious that we can’t make a square with side lesser than 1, because the frogs always have integer coordinates.