Problem:
Let R be the set of real numbers. Determine all functions f:R→R such that
f(x2−y2)=xf(x)−yf(y)
for all real numbers x and y.
Solution:
Setting x=y=0 in the given condition yields f(0)=0. Since
−xf(−x)−yf(y)=f[(−x)2−y2]=f(x2−y2)=xf(x)−yf(y)
we have f(x)=−f(x) for xî€ =0. Hence f(x) is odd. From now on, we assume xâ‹…y≥0. Setting y=0 in the given condition yields f(x2)=xf(x). Hence f(x2−y2)= f(x2)−f(y2), or, f(x2)=f(x2−y2)+f(y2). Since for x≥0 there is a muqum 1≥0 such that t2=x, it follows that
f(x)=f(x−y)+f(y)(11)
Setting x=2t and y=t in (1) gives
f(2t)=2f(t)
Setting x=t+1 and y=t in the given condition yields
f(2t+1)=(t+1)f(t+1)−tf(t)(3)
By (2) and by setting x=2t+1 and y=1 in (1), the left-hand side of (3) becomes
f(2t+1)=f(2t)+f(1)=2f(t)+f(1)(1)
On the other hand, by setting x=t+1 and y=1 in (1), the right-hand sile of 3 reads
(t+1)f(t+1)−tf(t)=(t+1)[f(t)+f(1)]−tf(t)=f(t)+(t+1)f(1)(151)
Putting (3), (4), and (5) together leads to 2f(t)+f(1)=f(t)+(t+1)f(1). or.
f(t)=tf(1)
for t≥0. Recall that f(x) is odd, we conclude that f(−i)=−f(t)=−tf(1) for t≥0. Hence f(x)=kx for all x, where k=f(1) is a constant. It is not difficult to see that all such functions indeed satisfy the conditions of the problem.
The problems on this page are the property of the MAA's American Mathematics Competitions