Proving G(x)=x||[x^2 mod 2^N] is not a PRG
Given as input a $2n$-bit string $x||y$, the distinguisher $D(x||y)$ outputs 1 if $y = x^2 \bmod{2^n}$ and $0$ otherwise.
If $x||y$ is uniformly distributed over $\{0,1\}^{2n}$, the probability that $y=x^2 \bmod{2^n}$ is $2^{-n}$. If $x||y$ is distributed as $G(X)$ where $X$ is uniform over $\{0,1\}^n$, the probability that $y=x^2 \bmod{2^n}$ is 1. The distinguishing advantage of $D$ is $1-2^{-n}$, which is non-negligible.