triviality of vector bundles with the reduced homology of base space entirely torsion Let $\xi$ be an $n$-dimensional vector bundle over a manifold $M$ such that the reduced cohomology 
$\tilde H^*(M;\mathbb{Z})$ is entirely torsion (every element has finite order under addition). 
Question: Prove that there exists an integer $k$ such that the $k$-fold Whitney sum $\xi^{\oplus k}$ is a trivial bundle.
How to solve this? My attempt: let the classifying map of $\xi$ be
$$
f: M\longrightarrow BO(n).
$$
The induced map on cohomology is
$$
f^*: H^*(BO(n);\mathbb{Z})=\mathbb{Z}[w,p]/I\longrightarrow H^*(M;\mathbb{Z})
$$
where $w$ is the Stiefel-Whitney classes and $p$ the Pontrjagin classes? (I am confused about this and did not find proper references). 
By the torsion property, there exists $k$ such that 
$$
w(\xi^{\oplus k})=p(\xi^{\oplus k})=1?
$$
Hence can this imply that $\xi$ is a trivial bundle?
Question: if the Stiefel-Whitney classes and Pontrjagin classes of a vector bundle $\eta$ are all trivial, can we obtain that $\eta$ is a trivial bundle?
 A: For Question 1 note first that it is important that the base space of the vector bundle be finite-dimensional, as one can see by looking at the canonical line bundle $\gamma$ over ${\mathbb R}P^\infty$ since $w_k(\gamma^{\oplus k})$ is nonzero for all $k\geq 1$, so $\gamma^{\oplus k}$ is not even stably trivial.
On the other hand, for a vector bundle $\xi$ over a base space $X$ which is a finite complex (CW or simplicial) with $H^i(X)$ finite for all $i>0$ then some sum $\xi^{\oplus k}$ is trivial.  This can be proved by an obstruction theory argument.  To begin, $\xi\oplus\xi$ is orientable, hence is trivial over the 1-skeleton of $X$.  Call this bundle $\xi_1$. The obstruction to $\xi_1$ being trivial over the 2-skeleton is $w_2$, at least if $\xi_1$ has dimension at least 3 which can be arranged by replacing the original $\xi$ by $\xi\oplus\xi$. Thus $\xi_2=\xi_1\oplus\xi_1$ will be trivial over the 2-skeleton, and hence also over the 3-skeleton since $\pi_2SO(n)=0$. Note that we haven't used finiteness of any of the groups $H^i(X)$ yet, but instead the finiteness of the first few homotopy groups of $SO(n)$.
Now we proceed inductively. Assume we have a bundle $\xi_n$ which is a sum of copies of $\xi$ and which is trivial over the $n$-skeleton of $X$. Choosing a trivialization over the $n$-skeleton, the obstruction to extending this trivialization over an $(n+1)$-cell is an element of $\pi_nSO(k)$ for $k$ the dimension of $\xi_n$. By obstruction theory, these elements define a cellular $(n+1)$-cochain on $X$ with coefficients in $\pi_nSO(k)$ which is in fact a cocycle. By rechoosing the framing on $n$-cells we can replace this cocycle by any other one in the same cohomology class. In particular, if the cohomology class is zero the bundle $\xi_n$ is trivial over the $(n+1)$-skeleton. This need not be the case initially, but since we assume that the cohomology groups of $X$ are finite, there is some multiple $m$ of the obstruction cocycle which is a coboundary. If we replace $\xi_n$ by $\xi_{n+1}=\xi_n^{\oplus m}$ then it is not hard to check that the obstruction cocycle for $\xi_{n+1}$ becomes zero in cohomology. (There are three ways to add classes in $\pi_n$ of orthogonal groups: (1) the usual addition in $\pi_n$, (2) by using the group structure in the orthogonal groups, and (3) by forming direct sums of matrices. These all coincide, after forming the direct sum in the third case.)  Thus $\xi_{n+1}$ is trivial over the $(n+1)$-skeleton, finishing the induction step when $X$ is a finite complex.  The example with ${\mathbb R}P^\infty$ shows that we may not be able to carry the induction through an infinite number of steps.  If $X$ were finite-dimensional but not finite, one would be able to finish if there were an upper bound on the orders of the elements of $H^i(X)$ for $i>0$.
A: Question 1: Stable triviality of some multiple $\xi^{\oplus k}$ follows under the given condition on homology from the Atiyah-Hirzebruch spectral sequence for K-theory, which gives that $\tilde{K}(X)$ is torsion.
Question 2: Vanishing of Pontryagin classes of $\xi$ shows stable triviality
of some multiple $\xi^{\oplus k}$: The reason is that stable vector bundles are given by maps to $BO$, and rationally $BO$ is a product of Eilenberg-MacLane spaces of dimension $4j$, corresponding to Pontryagin classes.
Finally for every stably trivial vector bundle $\eta$, some multiple $\eta^{\oplus k}$ is trivial. 
This is a theorem by Lam:
LAM, T.Y. Series summation of stably free modules,
Q J Math (1976) 27 (1): 37-46. doi: 10.1093/qmath/27.1.37
