= Solution
For a complex vector bundle $E$, define the kth <Adams operation> by the kth Newton polynomial in its exterior powers:
$$
\psi^k(E)=N_k(\Lambda^1E,\ldots,\Lambda^kE).
$$
After applying the <splitting principle for complex vector bundles> and writing $E=L_1\oplus\cdots\oplus L_r$, the <Newton identities> give
$$
\psi^k(E)=L_1^k+\cdots+L_r^k.
$$
This formula proves additivity. Applied simultaneously to splittings of $E$ and $F$, it also gives
$$
\psi^k(E\otimes F)=\sum_{i,j}(L_iM_j)^k
=\psi^k(E)\psi^k(F).
$$
Thus $\psi^k$ extends to a natural ring endomorphism of $K^0(X)$ and, for a line bundle,
$$
\boxed{\psi^k(L)=L^k.}
$$
Write $X=\mathbb{CP}^{k+2}_k$ and let $x=[\overline\gamma]-1$. Its reduced K-theory is freely generated by $x^k,x^{k+1},x^{k+2}$, with $x^{k+3}=0$, and restriction to the bottom cell sends $x^k$ to a <Bott element> $u\in\widetilde K^0(S^{2k})$ and the two higher powers to zero. If $r:X\to S^{2k}$ retracts the bottom-cell inclusion, then
$$
r^*(u)=x^k+ax^{k+1}+bx^{k+2}
$$
for some integers $a,b$. Naturality and the <Adams operation on a Bott class> give
$$
\psi^2(r^*u)=r^*(\psi^2u)=2^kr^*u.
$$
Since $\psi^2(x)=2x+x^2$, comparison modulo $x^{k+3}$ first in degree $k+1$ and then in degree $k+2$ gives
$$
a=-\frac{k}{2},
\qquad
b=\frac{k(3k+5)}{24}.
$$
The first equality makes $k$ even. Then $3k+5$ is odd, so integrality of $b$ forces $8\mid k$; it is also congruent to $2$ modulo $3$, so $3\mid k$. Consequently
$$
\boxed{24\mid k.}
$$
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