Problem:
Let Z be the set of integers. Find all functions f:Z→Z such that
xf(2f(y)−x)+y2f(2x−f(y))=xf(x)2​+f(yf(y))
for all x,y∈Z with xî€ =0.
Solution:
Let f be a solution of the problem. Let p be a prime. Since p divides f(p)2,p divides f(p) and so p divides pf(p)2​. Taking y=0 and x=p, we deduce that p divides f(0). As p is arbitrary, we must have f(0)=0. Next, take y=0 to obtain xf(−x)=xf(x)2​. Replacing x by −x, and combining the two relations yields f(x)=0 or f(x)=x2 for all x.
Suppose now that there exists x0â€‹î€ =0 such that f(x0​)=0. Taking y=x0​, we obtain xf(−x)+x02​f(2x)=xf(x)2​, yielding x02​f(2x)=0 for all x and so f vanishes on even numbers. Assume that there exists an odd number y0​ such that f(y0​)î€ =0, so f(y0​)=y02​. Taking y=y0​, we obtain
xf(2y02​−x)+y02​f(2x−y02​)=xf(x)2​+f(y03​)
Choosing x even, we deduce that y02​f(2x−y02​)=f(y03​). This forces f(y03​)=0, as otherwise we would have f(2x−y02​)=(2x−y02​)2 for all even x and so y02​(2x−y02​)2= f(y03​) for all such x, obviously impossible. Thus f(2x−y02​)=0 for all even numbers x, that is f vanishes on numbers of the form 4k+3. But since x2f(−x)=f(x)2,f also vanishes on all x such that −x≡−1(mod4), that is on 4Z+1. Thus f also vanishes on all odd numbers, contradicting the choice of y0​. Hence, if f is not the zero map, then f does not vanish outside 0 and so f(x)=x2 for all x.
In conclusion, f(x)=0 for all x∈Z and f(x)=x2 for all x∈Z are the only possible solutions. The first function clearly satisfies the given relation, while the second also satisfies the Sophie Germaine identity
x(2y2−x)2+y2(2x−y2)2=x3+y6
for all x,y∈Z.
OR
f(0)=0 : If f(0)î€ =0, set x=2f(0) to obtain
2(f(0))2=2f(0)(f(2f(0)))2​+f(0)
that is
2(f(0))2(2f(0)−1)=f(2f(0))2
But 2(2f(0)−1) cannot be a perfect square since it is of the form 4k+2. So f(0)=0.
This problem and the solutions were suggested by Titu Andreescu and Gabriel Dospinescu.
The problems on this page are the property of the MAA's American Mathematics Competitions