Adams operation on a Bott class

If $u\in {\tilde{K}}^0(S^{2m})$ is a Bott element generator, then

Let ${\mathbb{CP}}_k^{k+n}={\mathbb{CP}}^{k+n}/{\mathbb{CP}}^{k-1}$. If the bottom-cell inclusion $S^{2k}={\mathbb{CP}}_k^k\to {\mathbb{CP}}_k^{k+2}$ admits a retraction, then $24$ divides $k$. Writing $x=[\overline{\gamma }]-1$, a retracted Bott generator has the form $x^k+a{x}^{k+1}+b{x}^{k+2}$. Comparing it with its image under ${\psi }^2(x)=2x+x^2$ gives $a=-k/2$ and $b=k(3k+5)/24$; integrality forces both $3$ and $8$ to divide $k$.