Write $M\_n = S^n \cup\_2 D^{n+1}$. I know, as a matter of folklore, that the identity map $\mathrm{id}\_M$, considered as an element of the group $[M\_n, M\_n]$, has order $4$ (for $n > 3$, let's say). 


What is a reference for this? And is there a simple and pretty argument?



Order of the identity map of a Moore space.
Alternatively: if the order were $2$ then $M\_n\wedge M\_n$ would be $M\_{2n}\vee M\_{2n+1}$. The mod $2$ cohomology of $M\_n$ has a generator $a$ in degree $n$ and a generator $b=Sq^1(a)$ in degree $n+1$ and nothing else. It follows that the cohomology of $M\_n\wedge M\_n$ has generators $a\otimes a$, $a\otimes b$, $b\otimes a$ and $b\otimes b$ and we find using the Cartan formula that $Sq^2(a\otimes a)=b\otimes b$. However, $Sq^2$ is zero on the cohomology of $M\_{2n}\vee M\_{2n+1}$.


