You were right on the “or there should be some other criteria” part, but not on the “They need to be independent” part. If you start with independent Gaussians

You were right on the “or there should be some other criteria” part, but not on the “They need to be independent” part. If you start with independent Gaussians A and B and accept that A+B and A-B are Gaussians you’ll easily see that the sum of those is also Gaussian (because it’s 2A) even though they are not independent.