Variance of sum equals sum of variances is not sufficient for pairwise uncorrelated for more than 2 random variables. Simplest counterargument is any Cov(A,B)

Variance of sum equals sum of variances is not sufficient for pairwise uncorrelated for more than 2 random variables. Simplest counterargument is any Cov(A,B)=-Cov(B,C)!=0 for three variables. Closest results I can think of to the "iff" they're getting at: If sum var=var sum for all linear transformations of the individual variables (i.e. X_i -> a_i*X_i), that's sufficient for pairwise uncorrelated; If sum var=var sum for all transformations of the variables for which variances exist (i.e. X_i -> f_i(X_i)), that should be sufficient for independence (but I don't think that's an easy proof and maybe I'm missing technical conditions).