Problem:
Find all functions f:R→R such that for all real numbers x and y,
(f(x)+xy)⋅f(x−3y)+(f(y)+xy)⋅f(3x−y)=(f(x+y))2
Solution:
First, taking x=y=0 in the given yields f(0)=0, and then taking x=0 gives f(y)f(−y)=f(y)2. So also f(−y)2=f(y)f(−y), from which we conclude f is even. Then taking x=−y gives
∀x∈R:f(x)=x2 or f(4x)=0
for all x.
Next, we claim that
∀x∈R:f(x)=x2 or f(x)=0()
To see this assume f(t)î€ =0 (hence tî€ =0 ). By (⋆) we get f(t/4)=t2/16. Now take (x,y)=(3t/4,t/4) to get
4t2​f(2t)=f(t2)⟹f(2t)î€ =0
If we apply (⋆) again we actually also get f(t/2)î€ =0. Together these imply
f(t)î€ =0⟺f(2t)î€ =0
Repeat () to get f(4t)î€ =0, hence f(t)=t2, proving (Ω).
We are now ready to show the claimed solutions are the only ones. Assume there's an aî€ =0 for which f(a)=0; we show that f≡0. There are two approaches from here, by using inequalities or polynomials.
First approach
Pick b∈R, we show directly f(b)=0.
First, note that f≥0 always holds by ( ◯ ). By using ( 0 ) we can generate c>100b such that f(c)=0 (by taking c=2na for n large). Now, select x,y>0 such that x−3y=b and x+y=c id est
(x,y)=(43c+b​,4c−b​)
Substitution into the original equation gives
0=(f(x)+xy)f(b)+(f(y)+xy)f(3x−y)
But everything on the right-hand side is nonnegative. Thus it follows that f(b)= f(3x−y)=0 as desired.
Secod approach
First, observe that for all x \in \mathbb
f(4x−a)î€ =0⟹(f(x)+x(3x−a))f(3a−8x)=f(4x−a)2î€ =0
by taking y=3x−a in the original equation. Finally, consider the equations
​0=(4x−a)4−(x(3x−a))(3a−8x)20=(4x−a)4−(x2+x(3x−a))(3a−8x)2​
Each right-hand side is a nonzero polynomial in x. Thus there are finitely many roots in x, hence there are only finitely many values of x with f(4x−a)î€ =0. But ( Q) then implies there cannot be any values of x at all, i.e. we conclude that f≡0.
The problems on this page are the property of the MAA's American Mathematics Competitions