Bmo | 2008 Solutions ~upd~

Try linear solutions: f(x)=kx. Then f(f(x))=k^2 x = x ⇒ k=±1. Check k=1: LHS f(x f(y)+f(x)) = x y + x, RHS y x + x works. k=-1: f(x f(y)+f(x)) = -( -x y - x )? Let’s check: f(x f(y)+f(x)) = -[ x(-y) + (-x) ] = -[ -xy -x] = xy+x. RHS= y(-x)+x = -xy+x. Not equal unless x=0. So only f(x)=x works? But BMO 2008 solutions often include f(x)=-x? Let’s test: f(x)=-x: LHS= -[ x(-y) + (-x) ] = -[ -xy - x] = xy+x. RHS= y(-x) + x = -xy+x. So xy+x = -xy+x ⇒ 2xy=0 for all x,y ⇒ no. So only f(x)=x. But is that the only? Yes, after proving f additive, etc.

Solving a system of three equations for real values of bmo 2008 solutions

The correct : Use the fact that there are 8 black squares. Place the 8 smallest numbers (1-8) on black squares? Then white squares have 9-16. Any adjacent pair has at least one white and one black (chessboard), so difference at least 9-8=1, not 9. But we need at least 9. So we need a stronger invariant. Try linear solutions: f(x)=kx