Self-dual condition implies dimension 2 modulo 4

Question

Reading the article of Henneaux and Teitelboim Dynamics of Chiral (Self-Dual) p-Forms they state that "In order for self-dual fields to exist it is necessary that $F$ and $^*F$ should have the same number of components. It is also necessary that the square of the operation of taking the dual should give $+1$. These two demands restrict the spacetime dimension to be equal to $2$ modulo $4$".

My attempt was the following. If $A$ is a $p$-form, then $F=dA$ is a $(p+1)$-form of the form \begin{align} F=\frac{1}{(p+1)!}F_{\mu_1\mu_2\dots\mu_{p+1}}dx^{\mu_1}\wedge\dots\wedge dx^{\mu_{p+1}}\,. \end{align} Its Hodge dual is \begin{align} ^*F=\frac{\sqrt{|g|}}{(p+1)!(n-(p+1))!}F_{\mu_1\dots\mu_{p+1}}\epsilon^{\mu_1\dots\mu_{p+1}}{}_{\mu_{p+2}\dots \mu_n}dx^{\mu_{p+2}}\wedge \dots\wedge dx^{\mu_n}\,. \end{align} Equating both sides, one realize that the same number of components implies equation $p=n-(p+2)$, which yields $n=2(p+2)$. On the other side, on a Lorentzian manifold, the square of the Hodge operator is ${^*}{^*}F=(-1)^{1+(p+1)(n-(p+1))}F$. Taking this operation equals to $+1$, we arrive to the condition \begin{align} 1+(p+1)(n-(p+1))=2k\,, \end{align} with $k=0,1,2,3\dots$. Solving this condition implies that $n=\sqrt{8k}$?

How can I get to the condition that the authors say?

Answer 1

It's somewhat simpler to frame this in terms of $q=p+1$, in which case your conditions become that $q=n-q$ and that $q(n-q)$ is odd. The former condition implies that $n=2q$ must be even, while the latter implies that $q$ must be odd; the desired result follows immediately.