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
We claim that the only two functions satisfying the requirements are f(x)≡0 and f(x)≡x2. These work.
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)=x2orf(4x)=0(★)
for all x.
Remark Note that an example of a function satisfying (★) is
f(x)=⎩⎪⎪⎨⎪⎪⎧x21−cos(2π⋅x1337)0if ∣x∣<1if 1≤∣x∣<4if ∣x∣≥4.
So, yes, we are currently in a world of trouble, still. (This function is even continuous; I bring this up to emphasize that "continuity" is completely unrelated to the issue at hand.)
Now we claim
Claim f(z)=0⟺f(2z)=0(♠).
Proof. Let (x,y)=(3t,t) in the given to get
(f(t)+3t2)f(8t)=f(4t)2.
Now if f(4t)=0 (in particular, t=0), then f(8t)=0. Thus we have (♠) in the reverse direction.
Then f(4t)=0⟹(★)f(t)=t2=0⟹(♠)f(2t)=0 implies the forwards direction,the last step being the reverse direction (♠).
By putting together (★) and (♠) we finally get
f(x)=x2orf(x)=0(♡)
We are now ready to approach the main problem. Assume there's an a=0 for which f(a)=0; we show that f≡0.
Let b∈R be given. Since f is even, we can assume without loss of generality that a,b>0. Also, note that f(x)≥0 for all x by (♡). By using (♠) we can generate c>b such that f(c)=0 by taking c=2na for a large enough integer n. Now, select x,y>0 such that x−3y=b and x+y=c. That is,
(x,y)=(43c+b,4c−b).
Substitution into the original equation gives
0=(f(x)+xy)f(b)+(f(y)+xy)f(3x−y)≥(f(x)+xy)f(b).
But since f(b)≥0, it follows f(b)=0, as desired.
The problems on this page are the property of the MAA's American Mathematics Competitions