Difficult question 2

這條一定要 f(m+n) - f(m) - f(n) = 0 or 1 先做到. 數學家直覺都係話題目打錯. (ten13kidney 都指出了矛盾地方)
f(2) = 0
f(2) = f(1) + f(1) --> f(1) = 0
f(3) = f(2) + f(1) = 0 or 1 --> f(3) = 1
now notice: f(n+3) = f(n) + f(3) + 0/1 = f(n) + 1/2, so 每3最少進一
and f(n+2) = f(n) + f(2) = f(n) + 0/1, similarly f(n+1) = f(n) + 0/1, so 每格最多進一
so we have 3333 = f(9999) >= f(9996) + 1 >=... >= f(3) + 3332 = 3333, 所以一定有 f(3n) = n 是個等號, 亦即是說, 每3最多進一.
而我地有 f(4) = f(2)+f(2) + 0/1 = 1
如果f(5) = 1, f(6) = 2, 於是 f(3n) = f(3n+1) = f(3n+2) = n, 於是 f(1982) = 1980/3 = 660
如果f(5) = 2, f(6) = f(7) = 2, 於是 f(3n-1) = f(3n) = f(3n+1) = n, 於是 f(1982) = 661
邊個先對?
留意如果係後者, 將會有 f(10) = 3. 但 f(10) = f(5)+f(5) + 0/1 = 4/5, 不可能.
於是答案係 f(1982) = 660.

好以有矛盾！
因為f(2)=0,所以 f(3)=f(3)-f(2)-f(2)= f(1+2)-f(2)-f(2)= o or 1. 因為f取非負整數值，f(3)=1.
因為 1 or 0 = f(1+3)-f(3)-f(3) = f(4)-1-1, 所以f(4)= 2 or 3. 但是因為1 or 0 = f(2+2) -f(2) -f(2) = f(4) -0-0 =f(4), 所以f(4)=1. 矛盾.
再看多一個例子, 1 or 0 = f(3+2)-f(2)-f(2) =f(5)-0-0=f(5), f(5)=1. 但是, 1 or 0 = f(2+3)- f(3)-f(3)= f(5)-2, f(5)=2 or 3.矛盾.
請補充，勿删除。