Given functional equation $f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1)$, show that $f(0,0)=0$ [closed]

Question

Let $f:\mathbb{Z}\times \mathbb{Z}\to \mathbb{R}$, such that for all $x,y$ $$f(x,y)=f(x,y+1)+f(x+1,y)+f(x-1,y)+f(x,y-1),$$ and if $m,n\in \mathbb{Z},mn\neq 0$,we have $f(2m,2n)=0$.

Show that $$f(0,0)=0.$$

Answer 1

If this is true then there is a "finitary" proof, i.e. you can use the relations to express $f(0,0)$ as a finite linear combination of the $f(2n,2m)$.

First, here is a way to rephrase the problem in terms only of even entries of $f$, i.e. how to find a relation between $f(x,y),f(x+2,y),f(x,y+2),\ldots$

Consider $R = \Bbb R[X,X^{-1},Y,Y^{-1}]$ and let $P = (X+Y+X^{-1}+Y^{-1}-1)$.

$R$ is the set whose elements are finite linear combinations of monomials of the form $X^{\pm n} Y^{\pm m}$, and you can rephrase the goal by asking if there is an element $Q \in R$ such that $Q \times P$ has a nonzero constant coefficient and whose other nonzero coefficients occur only for exponents $(2n,2m)$ with nonzero $nm$.

Since $Q \times P$ is in $\Bbb R(X^2,Y^2)$ it is invariant by the Galois group $G$ of $\Bbb R(X,Y)$ over $\Bbb R(X^2,Y^2)$. $G$ has order $4$ and is given by the automorphisms $(X,Y) \mapsto (\pm X, \pm Y)$.

This tells us that $Q \times P$ has to be of the form $Q'(X^2,Y^2) \times P'(X^2,Y^2)$ where $P'(X^2,Y^2) = Norm(P) = P(X,Y)P(-X,Y)P(X,-Y)P(-X,-Y) $

You can easily compute $P'(X^2,Y^2) = (X^4 + Y^4 + X^{-4} + Y^{-4}) - 2(X^2Y^2 + X^2Y^{-2} + X^{-2}Y^{-2} + X^{-2}Y^2) - 2(X^2+Y^2+X^{-2}+Y^{-2}) -3$.

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This gives you the relations between the even entries of $f$, and now with this new polynomial, the problem reduces to finding $Q'$ such that $Q' \times P'$ has a nonzero constant coefficient and a zero coefficient in front of $X^n$ and $Y^m$.

Start with $1 \times P'$. It has a nonzero constant coefficient (-3) so we just have to "push out" the $8$ extra unwanted nonzero coefficients.

To do this, first add some $-\frac 29(X+Y+X^{-1}+Y^{-1}) \times P'$. This should make the $4$ entries near the constant coefficient zero while still keeping the constant coefficient nonzero (it is $-11/9$).

Next, by symmetry reasons you can remove the $X^{\pm2},Y^{\pm2}$ coefficients by adding a suitable multiple of $(X^2Y^2+X^{-2}Y^{-2}) \times P'$ without touching the constant and neighbouring coefficients.

Finally you can remove the remaining nonzero coefficients of $X^n$ one at a time by adding multiples of $X^nY^2 \times P'$ and similarly for $Y^n$.

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Computing the resulting polynomial gives you an explicit relation between $f(0,0)$ and a finite linear combination of some $f(2n,2m)$ with $nm \neq 0$, as needed.

Answer 2

My GNU Maxima script below,

a) Can't find a unique solution if its size parameter $N$ satisfies $N \le 6$.

b) Is able to prove $f(0,0) = 0$ if its size parameter $N$ satisfies $N \ge 7$.

c) Is able to prove $f(0,0) = 0$ if its size parameter $N$ satisfies $N \ge 3$

and one changes the constraint from $(mn \neq 0)$ into $(m \neq 0 \vee n \neq 0)$.

display2d : false;

/* variable construction */
my_v(f,m,n) := block(
    [res:[]],

    if (m < 0) and (n < 0) then block(
        res : concat(f,"m",-m,"m",-n)
    ),
    if (m < 0) and (n >= 0) then block(    
        res : concat(f,"m",-m,"p",+n)
    ),
    if (m >= 0) and (n < 0) then block(
        res : concat(f,"p",+m,"m",-n)
    ),
    if (m >= 0) and (n >= 0) then block(
        res : concat(f,"p",+m,"p",+n)
    ),

    return(res)
);

N : 7;
eqs : [];
for m : -N thru +N do block(
for n : -N thru +N do block(
    if (m*n # 0) then block(
        eqs : cons(my_v(f,2*m,2*n)=0,eqs)
    ),
    eqs : cons(my_v(f,m,n)
                = my_v(f,m,n-1) + my_v(f,m,n+1)
                    + my_v(f,m-1,n) + my_v(f,m+1,n), eqs)
));
print(eqs);
sols : algsys(eqs,listofvars(eqs));
s : sols[1];
for k thru length(s) do block(
    if listofvars(rhs(s[k])) = [] then block(
        print(k,"|",s[k])
    )
);

subst(s,fp0p0);

Switching off the generation of $f(2m,2n)=0$ in the above script and many variable eliminations later,

I have the following distilled equation, still symmetric in the arguments of f

$$ 46 f(+0,+0)\\ =\\ +13 f(-6,-2) -26 f(-4,-4) +13 f(-2,-6)\\ +13 f(+6,+2) -26 f(+4,+4) +13 f(+2,+6)\\ -6 f(-8,+2) -20 f(-4,-2) -20 f(-2,-4) -6 f(+2,-8)\\ -6 f(-2,+8) -20 f(+2,+4) -20 f(+4,+2) -6 f(+8,-2)\\ +24 f(-6,+2) +5 f(-2,-2) +24 f(+2,-6)\\ +24 f(-2,+6) +5 f(+2,+2) +24 f(+6,-2)\\ +6 f(-6,+4) +18 f(-4,+2) +18 f(-2,+4) +6 f(+4,-6)\\ +6 f(-4,+6) +18 f(+2,-4) +18 f(+4,-2) +6 f(+6,-4)\\ +34 f(-2,+2) +34 f(+2,-2). $$

from which the result follows.$\Box$

So this was somehow an algorithmic proof, the mathematical one follows now.

Additional symmetrization of the last equation gives $$ 92 f(+0,+0)\\ =\\ -6 f(+8,+2) +6 f(+6,+4) +6 f(+4,+6) -6 f(+2,+8)\\ -6 f(-8,-2) +6 f(-6,-4) +6 f(-4,-6) -6 f(-2,-8)\\ +37 f(+6,+2) -26 f(+4,+4) +37 f(+2,+6)\\ +37 f(-6,-2) -26 f(-4,-4) +37 f(-2,-6)\\ -6 f(+8,-2) -2 f(+4,+2) -2 f(+2,+4) -6 f(-2,+8)\\ -6 f(-8,+2) -2 f(-2,-4) -2 f(-4,-2) -6 f(+2,-8)\\ +37 f(+6,-2) +39 f(+2,+2) +37 f(-2,+6)\\ +37 f(+2,-6) +39 f(-2,-2) +37 f(-6,+2)\\ +6 f(+6,-4) -2 f(+4,-2) -2 f(-2,+4) +6 f(-4,+6)\\ +6 f(+4,-6) -2 f(+2,-4) -2 f(-4,+2) +6 f(-6,+4)\\ -26 f(+4,-4) +39 f(+2,-2) +39 f(-2,+2) -26 f(-4,+4). $$ Which can be reordered to $$ 92 f(+0,+0)\\ =\\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -6 f(8\sigma_x,2\sigma_y) +6 f(6\sigma_x,4\sigma_y) +6 f(4\sigma_x,6\sigma_y) -6 f(2\sigma_x,8\sigma_y)\\ +37 f(6\sigma_x,2\sigma_y) -26 f(4\sigma_x,4\sigma_y) +37 f(2\sigma_x,6\sigma_y)\\ -2 f(4\sigma_x,2\sigma_y) -2 f(2\sigma_x,4\sigma_y)\\ +39 f(2\sigma_x,2\sigma_y)). $$

This shall be our inspiration for the following definition $$ I(a,b,c,d)\\ := \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -6 f(8\sigma_x,2\sigma_y) +6 f(6\sigma_x,4\sigma_y) +6 f(4\sigma_x,6\sigma_y) -6 f(2\sigma_x,8\sigma_y)\\ +(c+37) f(6\sigma_x,2\sigma_y) +(d-26) f(4\sigma_x,4\sigma_y) +(c+37) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

On the following pages!! we will set the parameters $a,b,c,d$ as needed.

Removing the degree 10 terms gives $$ I(a,b,c,d)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -6 f(7\sigma_x,2\sigma_y) +6 f(7\sigma_x,3\sigma_y) +6 f(7\sigma_x,1\sigma_y)\\ -6 f(2\sigma_x,7\sigma_y) +6 f(3\sigma_x,7\sigma_y) +6 f(1\sigma_x,7\sigma_y)\\ +6 f(6\sigma_x,4\sigma_y) +6 f(4\sigma_x,6\sigma_y)\\ +(c+43) f(6\sigma_x,2\sigma_y) +(d-26) f(4\sigma_x,4\sigma_y) +(c+43) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +6 f(6\sigma_x,3\sigma_y) -6 f(5\sigma_x,3\sigma_y)\\ +6 f(3\sigma_x,6\sigma_y) -6 f(3\sigma_x,5\sigma_y)\\ -6 f(7\sigma_x,2\sigma_y) +6 f(7\sigma_x,1\sigma_y)\\ -6 f(2\sigma_x,7\sigma_y) +6 f(1\sigma_x,7\sigma_y)\\ +(c+37) f(6\sigma_x,2\sigma_y) +(d-26) f(4\sigma_x,4\sigma_y) +(c+37) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

Removing the degree 9 terms gives $$ I(a,b,c,d)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +6 f(5\sigma_x,2\sigma_y) +6 f(6\sigma_x,1\sigma_y)\\ +6 f(2\sigma_x,5\sigma_y) +6 f(1\sigma_x,6\sigma_y)\\ +12 f(6\sigma_x,3\sigma_y) -6 f(5\sigma_x,3\sigma_y)\\ +12 f(3\sigma_x,6\sigma_y) -6 f(3\sigma_x,5\sigma_y)\\ +6 f(7\sigma_x,1\sigma_y) +6 f(1\sigma_x,7\sigma_y)\\ +(c+31) f(6\sigma_x,2\sigma_y) +(d-26) f(4\sigma_x,4\sigma_y) +(c+31) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -12 f(4\sigma_x,3\sigma_y) -12 f(5\sigma_x,4\sigma_y)\\ -12 f(3\sigma_x,4\sigma_y) -12 f(4\sigma_x,5\sigma_y)\\ -6 f(5\sigma_x,2\sigma_y) +6 f(6\sigma_x,1\sigma_y)\\ -6 f(2\sigma_x,5\sigma_y) +6 f(1\sigma_x,6\sigma_y)\\ +6 f(5\sigma_x,3\sigma_y) +6 f(3\sigma_x,5\sigma_y)\\ +6 f(7\sigma_x,1\sigma_y) +6 f(1\sigma_x,7\sigma_y)\\ +(c+31) f(6\sigma_x,2\sigma_y) +(d-26) f(4\sigma_x,4\sigma_y) +(c+31) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -6 f(5\sigma_x,2\sigma_y) +6 f(6\sigma_x,1\sigma_y)\\ -6 f(2\sigma_x,5\sigma_y) +6 f(1\sigma_x,6\sigma_y)\\ +6 f(5\sigma_x,3\sigma_y) +6 f(3\sigma_x,5\sigma_y)\\ +6 f(7\sigma_x,1\sigma_y) +6 f(1\sigma_x,7\sigma_y)\\ +(c+31) f(6\sigma_x,2\sigma_y) +(d-38) f(4\sigma_x,4\sigma_y) +(c+31) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

The idea is now to remove degree by degree until nothing of positive degree is left.

Removing the degree 8 terms gives $$ I(a,b,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -6 f(5\sigma_x,1\sigma_y) -6 f(6\sigma_x,0\sigma_y)\\ -6 f(1\sigma_x,5\sigma_y) -6 f(0\sigma_x,6\sigma_y)\\ -6 f(5\sigma_x,2\sigma_y) +12 f(6\sigma_x,1\sigma_y)\\ -6 f(2\sigma_x,5\sigma_y) +12 f(1\sigma_x,6\sigma_y)\\ +6 f(5\sigma_x,3\sigma_y) +6 f(3\sigma_x,5\sigma_y)\\ +(c+25) f(6\sigma_x,2\sigma_y) -(2c+38) f(4\sigma_x,4\sigma_y) +(c+25) f(2\sigma_x,6\sigma_y)\\ +(b-2) f(4\sigma_x,2\sigma_y) +(b-2) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -(c+31) f(5\sigma_x,1\sigma_y) -6 f(6\sigma_x,0\sigma_y)\\ -(c+31) f(1\sigma_x,5\sigma_y) -6 f(0\sigma_x,6\sigma_y)\\ +(c+19) f(5\sigma_x,2\sigma_y) +12 f(6\sigma_x,1\sigma_y)\\ +(c+19) f(2\sigma_x,5\sigma_y) +12 f(1\sigma_x,6\sigma_y)\\ -(c+19) f(5\sigma_x,3\sigma_y) -(2c+38) f(4\sigma_x,4\sigma_y) -(c+19) f(3\sigma_x,5\sigma_y)\\ +(b-c-27) f(4\sigma_x,2\sigma_y) +(b-c-27) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -(c+19) f(4\sigma_x,3\sigma_y) -(c+19) f(3\sigma_x,4\sigma_y)\\ +(2c+38) f(3\sigma_x,3\sigma_y)\\ -(c+31) f(5\sigma_x,1\sigma_y) -6 f(6\sigma_x,0\sigma_y)\\ -(c+31) f(1\sigma_x,5\sigma_y) -6 f(0\sigma_x,6\sigma_y)\\ +(c+19) f(5\sigma_x,2\sigma_y) +12 f(6\sigma_x,1\sigma_y)\\ +(c+19) f(2\sigma_x,5\sigma_y) +12 f(1\sigma_x,6\sigma_y)\\ +(b-8) f(4\sigma_x,2\sigma_y) +(b-8) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

Removing the degree 7 terms gives $$ I(a,b,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -12 f(4\sigma_x,1\sigma_y) -12 f(5\sigma_x,0\sigma_y)\\ -12 f(1\sigma_x,4\sigma_y) -12 f(0\sigma_x,5\sigma_y)\\ -(c+19) f(4\sigma_x,3\sigma_y) -(c+19) f(3\sigma_x,4\sigma_y)\\ +(2c+38) f(3\sigma_x,3\sigma_y)\\ -(c+19) f(5\sigma_x,1\sigma_y) -6 f(6\sigma_x,0\sigma_y)\\ -(c+19) f(1\sigma_x,5\sigma_y) -6 f(0\sigma_x,6\sigma_y)\\ +(c+7) f(5\sigma_x,2\sigma_y) +(c+7) f(2\sigma_x,5\sigma_y)\\ +(b-8) f(4\sigma_x,2\sigma_y) +(b-8) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -(c+7) f(3\sigma_x,2\sigma_y) -(c+7) f(2\sigma_x,3\sigma_y)\\ -(c+19) f(4\sigma_x,1\sigma_y) -12 f(5\sigma_x,0\sigma_y)\\ -(c+19) f(1\sigma_x,4\sigma_y) -12 f(0\sigma_x,5\sigma_y)\\ -(2c+26) f(4\sigma_x,3\sigma_y) -(2c+26) f(3\sigma_x,4\sigma_y)\\ +(2c+38) f(3\sigma_x,3\sigma_y)\\ -(c+19) f(5\sigma_x,1\sigma_y) -6 f(6\sigma_x,0\sigma_y)\\ -(c+19) f(1\sigma_x,5\sigma_y) -6 f(0\sigma_x,6\sigma_y)\\ +(b+c-1) f(4\sigma_x,2\sigma_y) +(b+c-1) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(c+19) f(3\sigma_x,2\sigma_y) +(c+19) f(2\sigma_x,3\sigma_y)\\ -(c+19) f(4\sigma_x,1\sigma_y) -12 f(5\sigma_x,0\sigma_y)\\ -(c+19) f(1\sigma_x,4\sigma_y) -12 f(0\sigma_x,5\sigma_y)\\ +12 f(3\sigma_x,3\sigma_y)\\ -(c+19) f(5\sigma_x,1\sigma_y) -6 f(6\sigma_x,0\sigma_y)\\ -(c+19) f(1\sigma_x,5\sigma_y) -6 f(0\sigma_x,6\sigma_y)\\ +(b+c-1) f(4\sigma_x,2\sigma_y) +(b+c-1) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

We need another intermediate step, which we call $(*)$. We have for $n\ge 3$ $$ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( f(0\sigma_x,n\sigma_y)+f(n\sigma_x,0\sigma_y) )\\ =\\ +f(-0,-n)+f(-0,+n)+f(+0,-n)+f(+0,+n)\\ +f(-n,-0)+f(-n,+0)+f(+n,-0)+f(+n,+0)\\ =\\ +2f(0,-n)+2f(0,+n)+2f(-n,0)+2f(+n,0)\\ =\\ +2f(0,-n+1)-2f(0,-n+2)-2f(-1,-n+1)-2f(+1,-n+1) +2f(0,+n-1)-2f(0,+n-2)-2f(-1,+n-1)-2f(+1,+n-1) +2f(-n+1,0)-2f(-n+2,0)-2f(-n+1,-1)-2f(-n+1,+1) +2f(+n-1,0)-2f(+n-2,0)-2f(+n-1,-1)-2f(+n-1,+1)\\ =\\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +f(0\sigma_x,(n-1)\sigma_y)+f((n-1)\sigma_x,0\sigma_y)\\ +f(0\sigma_x,(n-2)\sigma_y)+f((n-2)\sigma_x,0\sigma_y)\\ -2f(1\sigma_x,(n-1)\sigma_y)-2f((n-1)\sigma_x,1\sigma_y) ). $$

With the help of $(*)$ this gives $$ I(a,b,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -6 f(4\sigma_x,0\sigma_y) -6 f(0\sigma_x,4\sigma_y) +(c+19) f(3\sigma_x,2\sigma_y) +(c+19) f(2\sigma_x,3\sigma_y)\\ -(c+19) f(4\sigma_x,1\sigma_y) -18 f(5\sigma_x,0\sigma_y)\\ -(c+19) f(1\sigma_x,4\sigma_y) -18 f(0\sigma_x,5\sigma_y)\\ +12 f(3\sigma_x,3\sigma_y)\\ -(c+7) f(5\sigma_x,1\sigma_y) -(c+7) f(1\sigma_x,5\sigma_y)\\ +(b+c-1) f(4\sigma_x,2\sigma_y) +(b+c-1) f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

Removing the degree 6 terms gives $$ I(a,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(c+7) f(3\sigma_x,1\sigma_y) +(c+7) f(1\sigma_x,3\sigma_y)\\ +(c+1) f(4\sigma_x,0\sigma_y) +(c+1) f(0\sigma_x,4\sigma_y)\\ +(c+19) f(3\sigma_x,2\sigma_y) +(c+19) f(2\sigma_x,3\sigma_y)\\ -(2c+26) f(4\sigma_x,1\sigma_y) -18 f(5\sigma_x,0\sigma_y)\\ -(2c+26) f(1\sigma_x,4\sigma_y) -18 f(0\sigma_x,5\sigma_y)\\ +12 f(3\sigma_x,3\sigma_y)\\ +6 f(4\sigma_x,2\sigma_y) +6 f(2\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(c+1) f(3\sigma_x,1\sigma_y) -12 f(2\sigma_x,2\sigma_y) +(c+1) f(1\sigma_x,3\sigma_y)\\ +(c+1) f(4\sigma_x,0\sigma_y) +(c+1) f(0\sigma_x,4\sigma_y)\\ +(c+25) f(3\sigma_x,2\sigma_y) +(c+25) f(2\sigma_x,3\sigma_y)\\ -(2c+26) f(4\sigma_x,1\sigma_y) -18 f(5\sigma_x,0\sigma_y)\\ -(2c+26) f(1\sigma_x,4\sigma_y) -18 f(0\sigma_x,5\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

With the help of $(*)$ this gives $$ I(a,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -18 f(3\sigma_x,0\sigma_y) -18 f(0\sigma_x,3\sigma_y)\\ +(c+1) f(3\sigma_x,1\sigma_y) -12 f(2\sigma_x,2\sigma_y) +(c+1) f(1\sigma_x,3\sigma_y)\\ +(c-17) f(4\sigma_x,0\sigma_y) +(c-17) f(0\sigma_x,4\sigma_y)\\ +(c+25) f(3\sigma_x,2\sigma_y) +(c+25) f(2\sigma_x,3\sigma_y)\\ -(2c-10) f(4\sigma_x,1\sigma_y) -(2c-10) f(1\sigma_x,4\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y)). $$

Removing the degree 5 terms gives $$ I(a,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(2c-10) f(2\sigma_x,1\sigma_y) +(2c-10) f(1\sigma_x,2\sigma_y)\\ +(2c-28) f(3\sigma_x,0\sigma_y) +(2c-28) f(0\sigma_x,3\sigma_y)\\ -(c-11) f(3\sigma_x,1\sigma_y) -12 f(2\sigma_x,2\sigma_y) -(c-11) f(1\sigma_x,3\sigma_y)\\ +(c-17) f(4\sigma_x,0\sigma_y) +(c-17) f(0\sigma_x,4\sigma_y)\\ +(3c+15) f(3\sigma_x,2\sigma_y) +(3c+15) f(2\sigma_x,3\sigma_y)\\ +(a+39) f(2\sigma_x,2\sigma_y))\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( -(c+25) f(2\sigma_x,1\sigma_y) -(c+25) f(1\sigma_x,2\sigma_y)\\ +(2c-28) f(3\sigma_x,0\sigma_y) +(2c-28) f(0\sigma_x,3\sigma_y)\\ -(c-11) f(3\sigma_x,1\sigma_y) -(c-11) f(1\sigma_x,3\sigma_y)\\ +(c-17) f(4\sigma_x,0\sigma_y) +(c-17) f(0\sigma_x,4\sigma_y)\\ +(a+3c+42) f(2\sigma_x,2\sigma_y)). $$

With the help of $(*)$ this gives $$ I(a,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(c-17) f(2\sigma_x,0\sigma_y) +(c-17) f(0\sigma_x,2\sigma_y)\\ -(c+25) f(2\sigma_x,1\sigma_y) -(c+25) f(1\sigma_x,2\sigma_y)\\ +(3c-45) f(3\sigma_x,0\sigma_y) +(3c-45) f(0\sigma_x,3\sigma_y)\\ -(3c-45) f(3\sigma_x,1\sigma_y) -(3c-45) f(1\sigma_x,3\sigma_y)\\ +(a+3c+42) f(2\sigma_x,2\sigma_y)). $$

Removing the degree 4 terms gives $$ I(-9c+48,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(4c-62) f(2\sigma_x,0\sigma_y) +(6c-90) f(1\sigma_x,1\sigma_y) +(4c-62) f(0\sigma_x,2\sigma_y)\\ -(4c-20) f(2\sigma_x,1\sigma_y) -(4c-20) f(1\sigma_x,2\sigma_y)\\ +(3c-45) f(3\sigma_x,0\sigma_y) +(3c-45) f(0\sigma_x,3\sigma_y)). $$

With the help of $(*)$ this gives $$ I(-9c+48,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(3c-45) f(1\sigma_x,0\sigma_y) +(3c-45) f(0\sigma_x,1\sigma_y)\\ +(7c-107) f(2\sigma_x,0\sigma_y) +(6c-90) f(1\sigma_x,1\sigma_y) +(7c-107) f(0\sigma_x,2\sigma_y)\\ -(10c-110) f(2\sigma_x,1\sigma_y) -(10c-110) f(1\sigma_x,2\sigma_y)). $$

Removing the degree 3 terms gives $$ I(-9c+48,-2c,c,-2c)\\ = \\ \sum_{\sigma_x,\,\sigma_y\in\{-1,+1\}} ( +(13c-155) f(1\sigma_x,0\sigma_y) +(13c-155) f(0\sigma_x,1\sigma_y)\\ +(7c-107) f(2\sigma_x,0\sigma_y) -(4c-20) f(1\sigma_x,1\sigma_y) +(7c-107) f(0\sigma_x,2\sigma_y)). $$

We are reaching the end of the proof. We need to remove the sum sign $$ I(-9c+48,-2c,c,-2c)\\ = \\ ( +(26c-310) f(+1,+0) +(26c-310) f(-1,+0)\\ +(26c-310) f(+0,+1) +(26c-310) f(+0,-1)\\ +(14c-214) f(+2,+0) +(14c-214) f(-2,+0)\\ -(4c-20) f(+1,+1) -(4c-20) f(+1,-1) -(4c-20) f(-1,+1) -(4c-20) f(-1,-1)\\ +(14c-214) f(+0,+2) +(14c-214) f(+0,-2)). $$

Using the functional equation for the middle gives $$ I(-9c+48,-2c,c,-2c)\\ = \\ ( +(14c-214) f(+2,+0) +(14c-214) f(-2,+0)\\ -(4c-20) f(+1,+1) -(4c-20) f(+1,-1) -(4c-20) f(-1,+1) -(4c-20) f(-1,-1)\\ +(14c-214) f(+0,+2) +(14c-214) f(+0,-2)\\ +(26c-310) f(+0,+0)). $$

On the other hand we have $$ -3 x f(+0,+0)\\ =\\ -4 x f(+0,+0) + x f(+0,+0)\\ =\\ -4 x f(+0,+0)\\ + x f(+2,+0) + x f(-2,+0) + x f(+0,+2) + x f(+0,-2)\\ + 2x f(+1,+1) + 2x f(+1,-1) + 2x f(-1,+1) + 2x f(-1,-1)\\ +4 x f(+0,+0)\\ =\\ + x f(+2,+0) + x f(-2,+0) + x f(+0,+2) + x f(+0,-2)\\ + 2x f(+1,+1) + 2x f(+1,-1) + 2x f(-1,+1) + 2x f(-1,-1). $$

Comparing now the coefficients of the $f(\cdot,\cdot)$ terms we get $$ +(14c-214) = +x \\ -(4c-20) = +2x $$

Solving for $c,x$ we have $c=-51/4$ and $x=71/2$. Using these we finish the proof with $$ I(-9c+48,-2c,c,-2c)\\ = \\ ( +(14c-214) f(+2,+0) +(14c-214) f(-2,+0)\\ -(4c-20) f(+1,+1) -(4c-20) f(+1,-1) -(4c-20) f(-1,+1) -(4c-20) f(-1,-1)\\ +(14c-214) f(+0,+2) +(14c-214) f(+0,-2)\\ +(26c-310) f(+0,+0))\\ =\\ -3 x f(+0,+0) +(26c-310) f(+0,+0))\\ =\\ -213/2 f(+0,+0) - 1283/2 f(+0,+0)\\ = -748 f(+0,+0). $$ Thus we have written $f(0,0)$ as a linear combination of the known vanishing terms even without using the vanishing property. $\Box$

I hope there are no typos somewhere.

Answer 3

Sketching a solution since there is a lot of tedium required. This method finds all solutions $f$; perhaps there is a shortcut to show only $f(0, 0) = 0$?

1. Show that if $f$, $g$ are solutions, then $f + g$ is a solution as well. Easy from the condition.
2. Show that if$$f(3, 4) = f(4, 3) = f(3, 6) = f(6, 3) = 0,$$then $f(x, y) = 0$ for all $x$, $y$. To show this, notice that$$f(3, 3) + f(5, 3) = 0$$from $(x, y) = (4, 3)$ in the condition, and$$f(3, 3) + f(3, 5) = 0$$from $(x, y) = (3, 4)$ in the condition. So $f(3, 5) = f(5, 3)$. Then since$$f(5, 3) + f(5, 5) = f(5, 4)$$and $$f(3, 5) + f(5, 5) = f(4, 5),$$we have $f(5, 4) = f(4, 5)$. Finally,$$f(5, 4) + f(4, 5) = 0$$from $(x, y) = (4, 4)$ in the condition, so$$f(5, 4) = f(4, 5)=0.$$This logic is the hardest step needed; showing all the other values have to be $0$ is long, but each step is either repeating what we just did or applying the original condition where four of the terms are $0$.
3. Show that if we treat $f$ as a function on the plane, then the following $4 \times 6$ array is a solution when tessellated infinitely, aligned so the top left corner is $(0, 0)$. Here, $a$ is an arbitrary real.$$\begin{matrix} 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & a & a & 0 & -a & -a \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -a & -a & 0 & a & a\end{matrix}$$
4. Show that if $f(3, 4)$, $f(4, 3)$, $f(3, 6)$, $f(6, 3)$ are arbitrarily set to real numbers, then there exists a solution. This is easy using step 1 and various transposes/translations of step 3.
5. Using steps 1 and 2, we have that if $f(3, 4)$, $f(4, 3)$, $f(3, 6)$, $f(6, 3)$ are specified, then there is at most one solution $f$. This means that step 4 describes all possible solutions. These solutions all satisfy $f(2m, 2n) = 0$ for all $m$, $n$, so we are done.