= Solution
All <Topological K-theory> groups below are complex and are indexed modulo two using <Bott periodicity>. For $n\geq1$, <Complex K-theory of a sphere> gives
$$
\boxed{
\begin{array}{c|cc}
&K^0(S^n)&K^{-1}(S^n)\\ \hline
n\text{ even}&\mathbb Z\oplus\mathbb Z&0\\
n\text{ odd}&\mathbb Z&\mathbb Z
\end{array}}
$$
For an even-dimensional <sphere>, choose a generator $u$ of <Reduced topological K-theory>. Its square vanishes: the <Chern character> sends $u$ to a top-degree <cohomology> class, whose square is zero, and the <Chern character> is injective on this torsion-free <abelian group>. Thus the <ring> answers are
$$
\boxed{K^0(S^{2m})\cong\mathbb Z[u]/(u^2)\quad(m\geq1),\qquad K^0(S^{2m+1})\cong\mathbb Z.}
$$
The exceptional zero-dimensional <sphere> consists of two points: $K^0(S^0)=\mathbb Z\times\mathbb Z$ with coordinatewise multiplication and $K^{-1}(S^0)=0$. In particular its reduced generator is idempotent rather than square-zero.
For the <Euler characteristic> identity, induct over the cells of a <finite CW complex>. The starting <CW complex> consisting of $v$ vertices has $K^0\cong\mathbb Z^v$, $K^{-1}=0$, and $\chi=v$. If $X$ is obtained from $A$ by attaching one $d$-dimensional cell, the quotient is $S^d$, and the <K-theory six-term exact sequence> is
$$
\widetilde K^0(S^d)\longrightarrow K^0(X)\longrightarrow K^0(A)
\longrightarrow\widetilde K^{-1}(S^d)\longrightarrow K^{-1}(X)
\longrightarrow K^{-1}(A)\longrightarrow\widetilde K^0(S^d).
$$
These <abelian groups> are finitely generated by induction. Tensoring the <exact sequence> with $\mathbb Q$ preserves exactness. The alternating sum of dimensions in a cyclic six-term <exact sequence> is zero, so
$$
\begin{aligned}
\operatorname{rank}K^0(X)-\operatorname{rank}K^{-1}(X)
&=\operatorname{rank}K^0(A)-\operatorname{rank}K^{-1}(A)\\
&\quad+\operatorname{rank}\widetilde K^0(S^d)
-\operatorname{rank}\widetilde K^{-1}(S^d)\\
&=\chi(A)+(-1)^d=\chi(X).
\end{aligned}
$$
Therefore
$$
\boxed{\chi(X)=\operatorname{rank}K^0(X)-\operatorname{rank}K^{-1}(X).}
$$
There is a genuine convention issue in the nilpotence assertion. With the usual definition of <Reduced topological K-theory> as the kernel of restriction to one basepoint, \b[the assertion needs $X$ to be connected]. For $X=S^0$ based at its first point, $(0,1)$ lies in $\widetilde K^0(X)$ and satisfies $(0,1)^m=(0,1)$ for every $m\geq1$. Thus it is not a <nilpotent element>. For an arbitrary <finite CW complex>, the correct assertion uses the <rank map in topological K-theory> on every <connected component>:
$$
I(X)=\ker\bigl(K^0(X)\longrightarrow H^0(X;\mathbb Z)\bigr).
$$
For a <connected space>, $I(X)=\widetilde K^0(X)$.
Here is an induction proving the corrected assertion. On the vertices, $I(X)=0$. For a positive-dimensional cell attachment $A\subset X$, the ideal $J=\ker(K^0(X)\to K^0(A))$ is square-zero. Indeed, lift two elements of $J$ to <relative topological K-theory> $K^0(X,A)$. Their product is induced by the reduced diagonal
$$
X/A=S^d\longrightarrow S^d\wedge S^d=S^{2d}.
$$
This map is <null-homotopic>, since $d<2d$ and $S^{2d}$ is $(2d-1)$-connected. The <relative product in topological K-theory> is therefore zero, giving $J^2=0$. If $a\in I(X)$, its restriction belongs to $I(A)$, so induction gives $(a|_A)^m=0$ for some $m$. Then $a^m\in J$ and $a^{2m}=0$. We have proved \b[every <Topological K-theory> class of rank zero on every <connected component> is a <nilpotent element>], and hence the requested result for <connected> $X$.
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