I found a conjecture of four squares since two months ago. But I don't have a solution for the conjecture. This conjecture is nice result in Euclidean geometry. I hope that there is a solution:
Conjecture 1: Let $A_iB_iC_iD_i$ for $i=1,2,3,4$ be four squares in a plane, with $A_i \rightarrow B_i \rightarrow C_i \rightarrow D_i$ all counter clockwise, (or all clockwise) for $i=1,2,3,4$. Then show that:
$$Area(A_1A_2A_3A_4)+Area(C_1C_2C_3C_4)=Area(B_1B_2B_3B_4)+Area(D_1D_2D_3D_4)$$
Conjecture 2: Let $O_X$ be the centroids of $X_1X_2X_3X_4$ for $X=A,B,C,D$ then show that $O_AO_BO_CO_D$ be a square.
About the centroids conjecture: assume that $A_0 B_0 C_0 D_0$ is a unit square centered in the origin and embed the construction in the complex plane. Any vertex of $A_1 B_1 C_1 D_1$ is the image of the corresponding vertex in the $A_0 B_0 C_0 D_0$ square through the map:
$$ z \mapsto w_1\cdot z+t_1 \tag{1}$$ hence: $$ O_A = \frac{A_1+A_2+A_3+A_4}{4} = \frac{t_1+t_2+t_3+t_4}{4}+\frac{w_1+w_2+w_3+w_4}{4}\cdot A_0 \tag{2} $$ but obviously $B_0 = i A_0, C_0=i B_0, D_0=i C_0$, so if we translate $O_A O_B O_C O_D$ by $-\frac{t_1+t_2+t_3+t_4}{4}$, the new points fulfill the same relation, hence they are the vertices of a square.
We also have that the centre of $O_A O_B O_C O_D$ is the centroid of the centres of the original squares.
We may tackle the area conjecture through the same approach. If $O,A,B$ are three points in the Euclidean plane, the area of $AOB$ is given by $\frac{1}{2}\left|(A-O)\times(B-O)\right|$. If we embed the Euclidean plane into the complex plane and assume that $O$ is the origin, then the (oriented) area of $AOB$ is given by $\frac{1}{2}\text{Im}\left(\overline{A}\cdot B\right)$.