若定义在R上的函数f(x)满足f(m+n)=f(m)+2[f(n)]^2 ,其中m,n属于R,f(1)≠0 则f(100)=?

f(2) = f(1+1) = f(1) + 2f(1)^2
f(3) = f(2+1) = f(2) + 2f(1)^2 = f(1) + 4f(1)^2
f(3) = f(1+2) = f(1) + 2f(2)^2 = 8f(1)^4 + 8f(1)^3 + 2f(1)^2 + f(1)
8f(1)^4 + 8f(1)^3 + 2f(1)^2 + f(1) = f(1) + 4f(1)^2
4f(1)^2 + 4f(1) - 1 = 0,解得f(1) = -1±√2
2f(1)^2 = 3±2√2
所以f(100) = f(1) + 99*2f(1)^2 = 296 ± 197√2答案是 50 ...