Problem 5.
Find all functions f from the set R of real numbers into R which satisfy
for all x,y,z belonging to R the identity
f(f(x) + f(y) + f(z)) = f(f(x) - f(y)) + f(2xy + f(z)) + 2f(xz-yz).
1.
i) z = 0 代入，则由x,y对称性 f(f(x)-f(y)) = f(f(y)-f(x))
ii) z = 1 代入，则由1及 x,y对称性 f(x-y) = f(y-x)，故 f(a)=f(-a)
2.
y=0 代入
f(f(x) + f(0) + f(z)) = f(f(x) - f(0)) + f(f(z)) + 2f(xz).
知若 f(x1)=f(x2)，则 f(x1z) = f(x2z) for all z in R.
3.
z=1代入
f(f(x) + f(y) + f(1)) = f(f(x) - f(y)) + f(2xy + f(1)) + 2f(x-y)
及 y=-y, z = 1 代入，由1.
f(f(x) + f(y) + f(1)) = f(f(x) - f(y)) + f(-2xy + f(1)) + 2f(x+y)
两式相减知
f(x+y) - f(x-y) 只和 xy 有关，f(x+y)-f(x-y) = f(1+xy)-f(1-xy)
又由2. f(x+y)-f(x-y) = f(1) - f((1-xy)/(1+xy)), if xy=/=-1
4.
Claim：若f非常数，则f(u)=f(v) => u= +-v
反证法可设|u|>|v|，由2. f(ur)=f(vr) for all r in R.
可解方程 x+y = ur, x-y=vr，此时 xy = (u^2-v^2)r^2 /4 >0
故取 r= 2 sqrt(u^2-v^2)，则
0 = f(x+y)-f(x-y) = f(1) - f((1-xy)/(1+xy)) = f(1)-f(0)
再由2. f(x) = f(0) 为常数。
(Note: 易证满足方程之常数函数只有0)
5.
y=1/2, z=0 代入
f(f(x) + f(1/2) + f(0)) = f(f(x) - f(1/2)) + f(x + f(0))
x=1,-1 分别代入，因f(a)=f(-a)，故 f(f(0)+1)=f(f(0)-1)
故由4. f(0) +1 = +- (f(0) -1)
+不可能，取-，得f(0)=0
6.
x=y 代入
f(2f(x) + f(z)) = f(2x^2 + f(z)) (1)
又(1) z=0, 代入
f(2f(x)) = f(2x^2)
故由1. 2f(x) = 2x^2，或-2x^2，即 f(x)=x^2 或-x^2
(1) z=1, 代入
若f(z)=1
2f(x) + 1 = +- (2x^2+1)且 2f(x) = +- 2x^2，故 f(x)=x^2
若f(z)=-1
2f(x) - 1 = +- (2x^2-1), 2f(x) = +- 2x^2，故 f(x)=x^2
验算合，结论f(x) = 0 或 x^2