所以,-1
log2(n+1)(m+1)=0
(n+1)(m+1)=1
1+mn+m+n=1
m+n=-mn>0
f(m+n)-f(m^2)
=log2(m+n+1)-log2(m^2+1)
=log2(m+n+1)/(m^2+1)
=log2(1-mn)/(m^2+1)
>log2(1-mn)/(|m||n|+1)
=log2(1-mn)/(1-mn)
=log2 1
=0
所以,f(m^2)
=log2(m+n+1)-log2(n^2+1)
=log2(m+n+1)/(n^2+1)
=log2(1-mn)/(n^2+1)
=log2 1
=0
所以,f(m+n)