I don't know the rest of the problem statement, nor the result they arrive at, but consider this:

If you add two independent normal random variables, you just get a new normal variable (with its mean being the sum of the original means, and the variance being the sum of the original variances).

This derives from the convolution property of pdfs; if you you convolve two Gaussians, your convolution integral just happens to give you another instance of a Gaussian.

I don't know the rest of the problem statement, nor the result they arrive at, but consider this:

If you add two independent normal random variables, you just get a new normal variable (with its mean being the sum of the original means, and the variance being the sum of the original variances).

This derives from the convolution property of pdfs; if you you convolve two Gaussians, your convolution integral just happens to give you another instance of a Gaussian.

This also extends to the twodimensional case.

So, you can never tell apart the sum of two independent sources from one source, and there's uncountably many different ways of arriving at the same sum. Since the density is the same, so are all moments, and cumulants.

However, the problem statement mentions a fourth-order cumulant method, which I'm not familiar with. And it also does not claim the first and second normal normal random variables to be independent – the trick here might be that the result shows it can detect at least the presence of a Gaussian mixture here. In fact, the problem statement doesn't exactly state these were added together; I could also imagine, just based on the text excerpt, that it's 2048 samples, the first half of which have mean µ, and the second half mean of –µ.