This problem was proposed by Ilir Snopce.

Nice problem but a bit easy for BMO shortlist's last problem. We can easily prove the following by induction: Consider a rectangle $ABCD$ with side lengths $m,n$ where $m,n$ are positive integers and $AB=m$. Then if $m,n$ are both odd, $f(A)+f(B)+f(C)+f(D)=0$. If $m$ is odd and $n$ is even $f(A)+f(B)=f(C)+f(D)$ and finally if $m,n$ are both even , $f(A)+f(C)=f(B)+f(D)$. *** Now , let $N$ be a very large positive integer and consider a $N*N$ square and all lattice points in it. Then write the statement $N^2$ times for all the $N^2$ unit squares in it and see it as a system of $N^2$ linear equations and $(N+1)^2$ variables.Now consider all $5*5$ squares constructed with the right-angled triangles with side lengths $3,4,5$ (that is , the $5*5$ squares which have the sides with the angles $37,53$ with coordinate axes , where we are assuming that the $N*N$ square has sides parallel to coordinate axes). We can write the statement of the problem again by *** for these squares and add some new equations to the system with the same set of variables , but independent with all the pervious $N^2$ ones. We do the similar thing with $13*13$ squares($13^2=12^2+5^2$) and all such pairs which are side lengths of an right-angled triangle and all the new equations would be independent with the pervious ones.Clearly the number of new equations would exceed $2N+1$ and we have a system where the number of equations is more than variables and there's no constant term and only solution to the system is the trivial solution $0$. So the only solution to the problem is $f \equiv 0$ which works.

alinazarboland wrote: Nice problem but a bit easy for BMO shortlist's last problem. We can easily prove the following by induction: Consider a rectangle $ABCD$ with side lengths $m,n$ where $m,n$ are positive integers and $AB=m$. Then if $m,n$ are both odd, $f(A)+f(B)+f(C)+f(D)=0$. If $m$ is odd and $n$ is even $f(A)+f(B)=f(C)+f(D)$ and finally if $m,n$ are both even , $f(A)+f(C)=f(B)+f(D)$. *** Now , let $N$ be a very large positive integer and consider a $N*N$ square and all lattice points in it. Then write the statement $N^2$ times for all the $N^2$ unit squares in it and see it as a system of $N^2$ linear equations and $(N+1)^2$ variables.Now consider all $5*5$ squares constructed with the right-angled triangles with side lengths $3,4,5$ (that is , the $5*5$ squares which have the sides with the angles $37,53$ with coordinate axes , where we are assuming that the $N*N$ square has sides parallel to coordinate axes). We can write the statement of the problem again by *** for these squares and add some new equations to the system with the same set of variables , but independent with all the pervious $N^2$ ones. We do the similar thing with $13*13$ squares($13^2=12^2+5^2$) and all such pairs which are side lengths of an right-angled triangle and all the new equations would be independent with the pervious ones.Clearly the number of new equations would exceed $2N+1$ and we have a system where the number of equations is more than variables and there's no constant term and only solution to the system is the trivial solution $0$. So the only solution to the problem is $f \equiv 0$ which works. How do you know all those equations are independent?

There is a lot of ambiguity here and I don't think this is correct. Before saying anything conclusive, I have some questions: 1) You cannot construct a $5\times 5$ square only with $3,4,5$ right triangles, so what do you actually mean when you construct these $5\times 5$ squares? You can construct a $5\times 5$ square with four $3,4,5$ right triangles and by adding a unit square in the middle, but in that case you add some new points. The same question for the $13\times 13$ square. 2) How can you make sure that all the equations you add are independent? They appear not to be independent. The proof of (***) itself is obtained by manipulating linear combinations of the initial condition, applied on (some of) the unit squares which make up an $N\times N$ square. Thus, any equations that you can get by applying (***) can also be obtained from the initial equations for the unit squares, so the system isn't independent. Edit: @gvole when quoting large posts, please use a hide tag

alinazarboland wrote: I don't add new points. The $5*5$ squares which are constructed as you said , have all their vertices among the lattice points of the initial $N*N$ square. Okay, but some of the vertices of the $3,4,5$ right triangles you use in the construction are not lattice points. alinazarboland wrote: But those $5*5$ squares or $13*13$ squares have completely different nature , that is they're constructed in a complete independent way from the pervious ones and they don't have anything to do with any of the squares which have sides parallel to coordinate axes. Well you previously said that the vertices of these $5\times 5$ and $13\times 13$ squares are among the lattice points of the big $N\times N$ square. I'm trying to understand how you construct these squares but you won't tell me. alinazarboland wrote: but I appreciate if you guys prove they're not independent I have already explained why they are not independent. Well, at least I think I did, since I don't quite understand what exactly you're constructing and working with. But let me explain in detail what I said above. Look at the big $N\times N$ square. The equations given by the condition for each of the unit squares which make up this big $N\times N$ square give you a system. Then, if you use (***) on any other rectangle whose vertices are among the lattice points in the big $N\times N$ square, you get an equation which is a linear combination of some previous equations in the already existing system, so you add nothing new.

@above , the part you say that all the equations can be derived by the linear combination of the given $N^2$ equations is not quiet right I believe. Consider this :Consider the cartesian coordinates. Let $N=7$ and consider the points $(i,j)$ where $7 \geq i,j \geq 0$. And consider the $5*5$ square constructed by $(0,4),(3,0),(7,3),(4,7)$. The equation given by *** for this square , cannot be derived from the equations of the $1*1$ squares of the form $(x,y),(x+1,y),(x+1,y+1),(x,y+1)$ where $6 \geq x,y \geq 0$ (at least I wasn't able to present a linear combination maybe I'm wrong)

oVlad wrote: alinazarboland wrote: I don't add new points. The $5*5$ squares which are constructed as you said , have all their vertices among the lattice points of the initial $N*N$ square. Okay, but some of the vertices of the $3,4,5$ right triangles you use in the construction are not lattice points. Why does it matter which points he uses to construct when the final equation is true anyways and contains only lattice points?

alinazarboland wrote: @above , the part you say that all the equations can be derived by the linear combination of the given $N^2$ equations is not quiet right I believe. Consider this :Consider the cartesian coordinates. Let $N=7$ and consider the points $(i,j)$ where $7 \geq i,j \geq 0$. And consider the $5*5$ square constructed by $(0,4),(3,0),(7,3),(4,7)$. The equation given by *** for this square , cannot be derived from the equations of the $1*1$ squares of the form $(x,y),(x+1,y),(x+1,y+1),(x,y+1)$ where $6 \geq x,y \geq 0$ (at least I wasn't able to present a linear combination maybe I'm wrong) I never said lattice squares, I said unit squares which make up the $N\times N$ square. In your case, we are talking about a rotated lattice. The point is, regardless of what lattice you are working in, you can refrain to the systems induced by the unit squares. If you compare the systems from multiple lattices (which might overlap), then your idea might work, but you cannot just claim it works, you actually need a proof that those equations are still independent, which you lack. Complete_quadrilateral wrote: Why does it matter which points he uses to construct when the final equation is true anyways and contains only lattice points? Because it doesn't only contain lattice points. If you construct a lattice $5\times 5$ square with $3,4,5$ right triangles and a unit square in the middle, some vertices of these right triangles (the ones opposite to the hypothenuses) will not be lattice points.

oVlad wrote: Complete_quadrilateral wrote: Why does it matter which points he uses to construct when the final equation is true anyways and contains only lattice points? Because it doesn't only contain lattice points. If you construct a lattice $5\times 5$ square with $3,4,5$ right triangles and a unit square in the middle, some vertices of these right triangles (the ones opposite to the hypothenuses) will not be lattice points. alinazarboland wrote: @above , the part you say that all the equations can be derived by the linear combination of the given $N^2$ equations is not quiet right I believe. Consider this :Consider the cartesian coordinates. Let $N=7$ and consider the points $(i,j)$ where $7 \geq i,j \geq 0$. And consider the $5*5$ square constructed by $(0,4),(3,0),(7,3),(4,7)$. The equation given by *** for this square , cannot be derived from the equations of the $1*1$ squares of the form $(x,y),(x+1,y),(x+1,y+1),(x,y+1)$ where $6 \geq x,y \geq 0$ (at least I wasn't able to present a linear combination maybe I'm wrong) This is pretty much what I meant

@above in that scenario, the vertices opposite to the hypothenuses are not in the lattice induced by the $5\times 5$ square. At this point I feel like we're debating blindly (and quite frankly I lost the thread) so I will leave it up to @alinazarboland to fix their idea (although I am doubtful). I feel like whatever you do with (***), you won't get any new equations, everything just cancels out. But @alinazarboland, if you do find a formal proof, please post it.

@above Here's an example for $N=7$. The 4 bolded points are the desired $5*5$ squares. Now , every one of expressions $f(x,y)$ has appear in either 1,2 or 4 square and those numbers in squares are the number of times they need to be applied in the linear combination to get $f(0,4)+f(3,0)+f(7,3)+f(4,7)$. I started from the first diagonal of lattice points. Then the second diagonal , then the third and so on... And when checking the left-up most square , we need to apply it 1 time which is impossible for $(0,7)$ to be applied one time it should've been 0 times.

I'm glad that this problem sparked such a rich discussion. Here is the official solution by the author, pictures included as attachments:

steppewolf wrote: Here is the official solution by the author, pictures included as attachments: I really hope there exists another approach...

oVlad wrote: steppewolf wrote: Here is the official solution by the author, pictures included as attachments: I really hope there exists another approach... So do I, but we couldn't find another solution. However, the problem was considered to be risky to use in the contest mainly because it was somewhat similar to this problem.

steppewolf wrote: So do I, but we couldn't find another solution. However, the problem was considered to be risky to use in the contest mainly because it was somewhat similar to this problem. And yet it has nothing to do with that problem. This one is just vector bash... although with an elegant statement, I wouldn't find it suitable for a contest. Edit: @above, Nikola isn't the author. Also, I didn't say you were definitely wrong, but only that I think you are.

oVlad wrote: @above, Nikola isn't the author. Also, I didn't say you were definitely wrong, but only that I think you are. Yeah maybe...I also invite you and everyone else to give an example(or an existence proof) that for example for case $N=7$ those equations are linearly dependent with $f(0,4)+f(3,0)+f(7,3)+f(4,7)$.though I believe the way I prove the linearly independency works(See #17) ; at least for the case $N=7$...

We can also ask a similar questions in any number of dimensions. The general case is for a given $n $-dimensional polytope $P $, if a function $f:\mathbb{R}^n\mapsto\mathbb{R}$ has the property that for all sets $S $ of points that are vertices of a polytope $P'\sim P$, $\sum_{V\in S}f(V)=0$, must it be that $f\equiv 0$. Some examples where it is easily provable as a yes: 1. $n=2$ with $P $ a regular polygon. 2. Any dimension with $P $ a simplex. 3. $n=3$ with $P $ a cube.

@above the problem is easy if the condition holds for polytopes $P'{}$ similar with $P{}$. Did you mean to say congruent? If so, may you please post a proof?

For the simplex case, it even works if we say congruent, just like with the regular polygon case. The proof of the simplex case is by taking two congruent $n$-simplices which share a $(n-1)$-simplex cell. Taking the difference of the two corresponding equations gives $f(A)-f(B)=0$ where $A,B$ are the two unshared vertices. Thus $f $ is a constant function and applying the zero sum condition gives $f\equiv 0$. For the cube, the case with similar cubes is easy, since you can take a $3\times 3\times 3$ array of points consisting of vertices of a cube, midpoints of its edges, centers of its faces, and the cube centers. If you checkerboard-color the eight smaller cubes, we can sum the equations for four black cubes, and subtract the equations for the white cubes. Now subtract the resulting equation from the one for the large cube and divide by 2. The result is an equation that asserts the simplex case for the regular tetrahedron. This makes $f\equiv 0$.