# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Let $f$ be a real-valued function on the plane such that for every square $ABCD$ in the plane, $f(A)+f(B)+f(C)+f(D)=0.$ Does it follow that $f(P)=0$ for all points $P$ in the plane?

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.4" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]If the function is indeed identically 0, then you would need a geometrical argument to prove it. If not, you would need to find an example of a function which is not identically zero and satisfies the hypothesis of the question. In principle, you should start by playing around with some examples. [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]The answer to the question is YES. [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]Think about squares that have $P$ as a vertex. [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"] Consider the squares in picture. Clearly, $f(D)+f(C)+f(P)+f(E)=0 \\ f(C)+f(B)+f(I)+f(P)=0 \\ f(E)+f(P)+f(G)+f(F)=0 \\ f(P)+f(I)+f(H)+f(G)=0$ Adding, we get    $4f(P)+2(f(C)+f(E)+f(G)+f(I))+ (f(D)+f(B)+f(H)+f(F))=0$.   As both $CEGI$ and $DBHF$ are squares, the second and the third terms are zero. Thus, $f(P)=0$.   [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px"]