Prove geodesics are straight lines if Riemann tensor is identically zero. Suppose that $R^{a}_{bcd}\equiv 0$ in all of our manifold $M$ (in which we assume zero torsion). Prove that all geodesics are straight lines.
I tried using Ricci's identity: $X^i_{;jk}-X^i_{;kj}=R^i_{mkj}X^m$ but expanding the right side and equating to zero only led me to the expression for $R$ in terms of the connection coefficients $\Gamma$.
I was hinted that I might use the parallel transport property along a circuit or that in a torsionless setting, $\nabla_X\nabla_X Z=R(X,Z,X)=0$. My problem lies in the fact that I don't entirely understand that last equation, and most references on the subject make use of properties of a Riemannian metric on the manifold, concept not yet introduced at this point in my course.
All and any help would be appreciated.

The result is false. A quick class of counterexamples is the following: consider any diffeomorphism $f:\mathbb R^n \to \mathbb R^n$. Let $\delta$ be the Euclidean metric on $\mathbb R^n$. Then the pullback metric $f^*\delta$ is a flat metric on $\mathbb R^n$ in which the geodesics typically aren't straight lines.
I realize that this answer perhaps contains technical language that you haven't yet seen, but you will learn about metrics and pullback soon in your class (I hope!). Furthermore, the above paragraph is a point that seems to be missed by a lot of students, so keep it in mind for the future
:-)
