# Generation of the order $\lambda$ (which is lcm((p-1),(q-1))) element g in modified paillier, why $-a^{2n}$?

As the question states, in variants of paillier cryptosystem, such as CS01 and DT-PKC, when they want an element $$g$$ of order $$\lambda$$, they choose a random number $$a$$ from group $$Z^*_{n^2}$$ and calculate $$-a^{2n}$$ as $$g$$. First, what's this multiplication of $$-1$$ for? Second, why $$a^{2n}$$ not just $$a^{n}$$? I think $$-1$$ changes nothing and $$a^{2n}$$ will give us an element of order $$\lambda/2$$ more likely, not $$\lambda$$. Could anyone explain this for me? Thanks.

If we choose $$n$$ to be the product of two strong primes $$p=2r+1$$ and $$q=2s+1$$ with $$r$$ and $$s$$ prime, note that $$p$$ and $$q$$ are 3 mod 4 and that $$\mathrm{LCM}(p-1,q-1)=2rs$$. Choosing a random $$a$$ and raising it to the power $$2n$$ gives an element of order $$\lambda/2=rs$$ (there is a vanishingly small chance of getting order $$r$$, $$s$$ or $$1$$) and which is therefore a quadratic residue. Multiplying by -1 then makes it a non-residue and hence of order $$\lambda=2rs$$. It also ensures that the Jacobi symbol is 1 so that no information is leaked via such symbols.
If we did not do this, there would be a non-negligible chance that $$a$$ is a quadratic residue and hence that $$g$$ would be of order $$\lambda/2$$ rather than $$\lambda$$.