Proving that if $|G|=p^n$ then $\exists a \in G:|C(a)| = p^{n-1}$
Solution 1:
The Class Equation yields: $$p^n=p+\sum_{i=1}^k\frac{p^n}{p^{\alpha_i}}$$ where $1\le\alpha_i< n$ because $x_i$'s are noncentral (whence $\alpha_i<n$) and $Z(G)=\bigcap_{g\in G}C(g)$ (whence $1\le\alpha_i$). Therefore: $$p^{n-1}=1+\sum_{i=1}^kp^{n-\alpha_i-1}$$ Since $p$ divides the LHS (as $n\ge 2$), if $n-\alpha_i-1>0$ for every $i$ we have a contradiction, because then $p\mid \sum_{i=1}^kp^{n-\alpha_i-1}$ but $p\nmid 1$. So, necessarily $n-\alpha_i-1=0$ for some $i$, namely $\alpha_i=n-1$ for some $i$.