Fall 2015, problem 4

Hi, I'm Hubert from Paris; the nine following points contain six squares;
to simplify, we write $(a,b)$ instead of $f(a,b)$, with $x=(0,0)$.

$x+(0,1)+(-1,1)+(-1,0)=0$

$x+(1,0)+(1,1)+(0,1)=0$

$x+(1,0)+(1,-1)+(0,-1)=0$

$x+(0,-1)+(-1,-1)+(-1,0)=0$

$-2(0,-1)-2(1,0)-2(0,1)-2(-1,0)=0$

$-(-1,1)-(1,1)-(1,-1)-(-1,-1)=0$

adding these yields $4x=0$, hence $f\equiv 0$ by translation.

If $ p$ is a point in the plane, I will simply write $ p$, when I mean $ f (p) $. It should be clear from context.

Take any square $abcd$ and subdivide it into four smaller squares, thereby adding points $ efghi$. Schematically, you get nine points arranged something like this:

$ \begin{matrix} a & e & d\\ f & i & h\\ b & g & c \end{matrix} $

Sum up the corners of any of the small squares and you get zero, so:

$\begin{align} 4 \times 0 = &(a+f+i+e)+\\ &(f+b+g+i)+\\ &(i+g+c+h)+\\ &(e+i+h+d) \end{align}$

Now rearrange:

$\begin{align} 0= &4 i + (a + b+c + d) + \\ &2 (e + f+g +h) \end{align}$

Note that both brackets vanish, because $ abcd$ is a square and so is $ efgh$. Thus $ i=0$.

Since the point $ i$ is in no way special, it follows that $ p =0$ for every point $ p$.

Consider $v_1, v_2, v_3$ and $v_4$ in $\mathbb{R}^2$ the vertices of a square $S$, and such a function $f$; In addition let's name $M$ the centre of gravity of this $S$ in $\mathbb{R}^2$, and four other elements of $\mathbb{R}^2$ (points) $v_{12}$, $v_{23}$, $v_{34}$, $v_{41}$ where $v_ij$ is the midle of [v i vj]. Note that these last points are the vertices of another square $S_1$ in $\mathbb{R}^2$ and they also verify the equation of the problem, If we look at the four resulting squares, we have: $$ f(M)+f(v_{23})+f(v_3)+f(v_{34})=0 $$ $$ f(M)+f(v_{41})+f(v_4)+f(v_{34})=0 $$ $$ f(M)+f(v_{23})+f(v_2)+f(v_{12})=0 $$ $$ f(M)+f(v_{41})+f(v_1)+f(v_{12})=0 $$ Summing these equation, we get $$ 4f(_M)+f(v_1)+f(v_2)+f(v_3)+f(v_4)+2(f(v_{12})+f(v_{23})+f(v_{34})+f(v_{41}))=0 $$ Wich gives us $4f(M)=0$ so $f(M)=0$ For each $v_1,v_2,v_3,v_4$ and their centre of gravity in $\mathbb{R}^2$: $f(v_1)+f(v_2)+f(v_3)+f(v_4)=0 \Longrightarrow f(M)=0$ means that for each M in $\mathbb{R}^2$: $f(M)=0$, f is identically $0$.

• For one thing, if $v_1, v_1, v_1, v_1$ is a square, then $f$ must be identically $0$.

• If this is not allowed:

$\forall (m,n) \in \mathbb{R^2}$, let $f((m,n))=x$.

Take any positive $l \in \mathbb{R}$. Let $f((m,n+l))=a$, $f((m-l,n))=b$, $f((m+l,n))=c$. Then $f((m-l,n+l))=-a-b-x$ and $f((m+l,n+l))=-a-c-x$.

Since $(m,n+l)$, $(m-l,n)$, $(m+l,n)$, and $(m,n-l)$ form a square, $f((m,n-l))=-a-b-c$.

After filling in $f((m-l,n-l))=a+c-x$ and $f((m+l,n-l))=a+b-x$, because these two form a square with $f((m-l,n+l))=-a-b-x$ and $f((m+l,n+l))=-a-c-x$, we have -4x = 0, which means x=0. Thus, for all such $(m,n)$, $f((m,n))=x=0$.

@Nelix Thank you~~~~ ^_^

Let us consider the problem in the following stencil:

$$ f_1 \ f_2 \ f_3\\ f_4 \ f_5 \ f_6\\ f_7 \ f_8 \ f_9 $$

with the length 1. Then, we have 4 small squares with length 1. If we add the four equation together, we have

$ f_1+f_3+f_7+f_9+2(f_4+f_2+f_6+f_8)+4f_5=0 $

Notice that $f_1+f_3+f_7+f_9=0$ in the sense of the big square, we have

$ f_4+f_2+f_6+f_8+2f_5=0 $

That means

$ f_1+f_9=0 $

Next, we extend the square to

$$ f_1 \ f_2 \ f_3 \ \circ \ \circ \\ f_4 \ f_5 \ f_6 \ \circ \ \circ \\ f_7 \ f_8 \ f_9 \ f_{10} \ f_{11}\\ \circ \ \circ \ f_{12} \ f_{13} \ f_{14}\\ \circ \ \circ \ f_{15} f_{16} f_{17} $$

Then, by using the same argument we have $f_9+f_{17}=0$ and in the square with length 2, we have $f_1+f_{17}=0$. Combine with $f_1+f_9=0$, we have $f_1=f_9=f_{17}=0$.

Using this argument in the square with any length, we have $f=0$ in whole $\mathbb{R}^2$ plane.

PS: I do a apologize that I do not know how to inser picture here . /(ㄒoㄒ)/~~

Hi I'm Pavan, Consider the v1, v2, v3, v4 be the vertices of a closed four-sided square Consider ∑ j=1 to 4 f(v i )= v1 + v2 + v3 + v4 Adding – ( v1 + v2 + v3 + v4) to the summation can be taken as ∑ j=1 to 4 f(v i )= v1 + v2 + v3 + v4 – ( v1 + v2 + v3 + v4) The terms can be rearranged as ∑ j=1 to 4 f(v i )= v2 – v1 + v3 – v2 + v4 – v3 +v1 – v4 = v1v2 + v2v3 + v3v4 + v4v1 Where vectors v1v2, v2v3, v3v4, v4v1 are the sides of the square. This can be also be taken as the sum of the vector sides of a square is 0. Which can also be put as the sum vectors of the sides of the closed n-sided polygon is 0.

please let me know for the opinion and the answer to the problem.

Sketch of proof: The number of squares in a square lattice with side length n ( a positive integer) is just the sum of the squares from 1 to n. (This result is obtained by counting the number of squares in the lattice with side lengths 1, 2, ..., n and adding.) Label the points in this lattice (from top to bottom and left to right) as v11, v12, ..., v1n, v21, v22, ..., v2n, ... vn1, vn2, ..., vnn (as in a matrix.) Define aij = f(vij). A square lattice of side length n then has n^2 lattice points. Applying the "vertex values of f() on any square sum to zero" condition for the function f() on each square in the grid yields an equation for each such square in which four of the aij variables with coefficients equal to 1 are summed on the left-hand side, and the right-hand side is zero. Moreover, it should be clear that all of these equations are linearly independent. (This is the "sketch" part.) The number of squares in the lattice is (by the well-known formula) n( n+1)(2 n+1)/6, whereas the number of variables aij is just the number of lattice points, which is n^2. For n >= 4 the former is strictly greater than the latter, so that the number of equations exceeds the number of variables. In particular, if n = 4, we have a homogeneous non-singular system of 30 equations in 25 unknowns. The only possible solution is the trivial one, in which every aij = f(vij) = 0.

There are no smoothness constraints on f. Consequently, any solution of this problem must be based on linear algebra (using the relation/constraint provided) and the (square) lattice geometry of the plane with its independence of the equations on the distinct lattice squares' vertices. For the problem as stated, one could just write down the 30 x 25 sparse coefficient matrix and solve it explicitly for a complete (if tedious) proof.

Postscript: Hmmm, now that I think on it, I probably overstated the case a little. The essential idea is correct, however, you clearly cannot have 30 'independent' equations in 25 variables. However, I still believe that it is true that selecting any 25 of the 30 distinct squares in the 4 x 4 grid will yield 25 independent equations, and the result still follows.

For any given point $v_1$, let:

• $S_1$ have vertices $v_1, v_2, v_3, v_4$
• $S_2$ have vertices $v_1, v_2, v_5, v_6$
• $S_3$ have vertices $v_1, v_8, v_7, v_6$
• $S_4$ have vertices $v_1, v_8, v_9, v_4$

Then we can apply the given property that the sum of $f$ applied to the vertices of a square equals zero, giving us four equations. Adding them yields

$$4f(v_1) + 2f(v_2) + 2f(v_4) + 2f(v_6) + 2f(v_8) + f(v_3) + f(v_5) + f(v_7) + f(v_9) = 0$$

Since $v_2$, $v_4$, $v_6$, and $v_8$ are vertices of a square, and so are $v_3$, $v_5$, $v_7$, and $v_9$, we simplify the equation to

$$4f(v_1) = 0$$

or more simply,

$$f(v_1) = 0$$