A Is F F F B Is F F F C Is F F F: Fill & Download for Free

Just to illustrate Brian Sittinger’s solution (in case you’re too lazy to draw it)Let [math]I[/math] be a random point and draw [math]4[/math] identical squares [math]IFAH, IHDG,IFBE[/math] and [math]IECG[/math].The nice thing about this, is that [math]ABCD[/math] and [math]FHGE[/math] also turn out to be squares!By the problem statement we have:[math]f(I)+f(H)+f(A)+f(F)=0 \tag{1}[/math][math]f(I)+f(H)+f(D)+f(G)=0 \tag{2}[/math][math]f(I)+f(F)+f(B)+f(E)=0 \tag{3}[/math][math]f(I)+f(E)+f(C)+f(G)=0 \tag{4}[/math][math]f(H)+f(F)+f(E)+f(G)=0 \tag{5}[/math][math]f(A)+f(B)+f(C)+f(D)=0 \tag{6}[/math]Now:[math] 4f(I)=(1)+(2)+(3)+(4) - 2(5)-(6)=0\tag*{}[/math]Hence [math]f(I)=0[/math] and [math]f[/math][math][/math] is the zero function.

Since you have not specified the domain of the function, the simplest solution would be a domain of three arbitrary elements, say [math]\{a,b,c\}[/math], where:[math]\quad f(a)=b; f(b)=c; f(c)=a[/math]This can be extended to any domain [math]D\supset\{a,b,c\}[/math] by simply defining [math]\forall x\notin\{a,b,c\}\colon f(x)=x[/math]. That would work for functions of Real numbers with [math]\{1,2,3\}[/math] for example.If you also want the function to be continuous with a non-trivial topology, the simplest solution is probably the Complex numbers with [math]f\colon\C\to\C[/math] defined by[math]\quad f(z)=\omega z[/math] where [math]\omega=\frac{-1+i\sqrt3}{2}[/math] is a cube-root of unity.Complex [math]n^{\text{th}}[/math] roots of unity give you a solution for a non-identity continuous function such that [math]f^n(z)\equiv z[/math] for any [math]n\in\N[/math]. These can be thought of as the obvious rotations of the plane by [math]\frac{360^{\circ}}{n}[/math].There is no continuous Real function such that [math]f^3(x)\equiv x[/math] other than the identity function.

Suppose [math]f[/math][math][/math] is a polynomial. Then [math]f[/math][math][/math] must have degree 1 since [math]f(f(x))[/math] has degree 1. So, let’s see if there is a solution of the form [math]f(x)=ax+b[/math].Since [math]f[/math][math][/math] in this case is differentiable, applying the chain rule to [math]f(f(x))=1-x[/math] we get [math]f’(f(x))f’(x)=-1[/math]. Since [math]f’(x)=a[/math], this immediately gives [math]a^2=-1[/math]. If we require [math]a,b \in \mathbb{R}[/math] then no polynomial solution exists. But if we allow [math]a,b \in \mathbb{C}[/math] then there are two cases:With [math]a=i[/math], we get [math]f(f(x))=i(ix+b)+b=-x+b(1+i)[/math]. Thus, [math]b=(1-i)/2[/math], and so [math]f(x)=ix+(1-i)/2[/math] is one solution.With [math]a=-i[/math], we get [math]f(f(x))=-i(-ix+b)+b=-x+b(1-i)[/math]. Thus, [math]b=(1+i)/2[/math], and so [math]f(x)=-ix+(1+i)/2[/math] is another solution.These are the two complex-coefficient polynomial solutions. Are there other solutions?Ján Bábeľa's answer shows that there is no real continuous function as a solution.