parallel curvature imply constant Ricci and scalar curvature $\text{Suppose we have} \nabla R = 0 $, where R represents curvature tensor, Prove that Ricci  curvature and scalar curvature are constant. 
 A: I am assuming a
Riemannian manifold,
and I am going to use its
Levi-Civita connection.
Contrary to my initial assumption, the constant in the problem statement
is not just supposed to mean that the covariant derivative vanishes.
And you already know that covariant differentiation and contraction commute.
I apologize for having used such a simple-minded interpretation
in the comments.
To fix the wording, a tensor field with vanishing covariant derivative
is called parallel here, not constant.
In fact, as Ted Shifrin has pointed out, any 
locally Riemannian symmetric space
features a parallel Riemann curvature tensor field,
but this does not make it a space of constant curvature.
We only require a parallel Riemann curvature tensor field,
but seek to establish a constant Ricci tensor field
and a constant scalar curvature. The latter is no problem,
as for scalar fields, parallel is still the same as constant.
For the Ricci tensor field, interpreted as a family of bilinear forms,
being constant in the strong sense means
$$\exists\lambda\in\mathbb{R}: \forall p\in M:
\forall X\in \operatorname{T}_p M:
\operatorname{Ric}(X,X) = \lambda\langle X,X\rangle$$
That property characterizes
an Einstein manifold.
So you are essentially asking whether every
locally Riemannian symmetric space is an Einstein manifold.
The answer is: No, but you seem to be close.
An easy counterexample to the proposition in question is
$$M = S\times T\quad\text{where}\quad S=\mathbb{S}^2,\ T = \mathbb{R}$$
the Riemannian product of the unit sphere $S$ and a line $T$.
For $p\in M$, decompose $p=(s,t)$ where $s\in S$, $t\in T$.
Likewise in the tangent space, the composition of $M$ allows you to
assemble $X\in \operatorname{T}_p M$ as $X=(X_s,X_t)$
where $X_s\in \operatorname{T}_s S$ and $X_t\in \operatorname{T}_t T$.
Furthermore, both $(X_s,0)$ and $(0,X_t)$ are in $\operatorname{T}_p M$,
and we have
$$\left\langle(X_s,X_t),(X_s,X_t)\right\rangle_p =
\langle X_s,X_s\rangle_s + \langle X_t,X_t\rangle_t$$
where the inner products are those of $\operatorname{T}_p M$,
$\operatorname{T}_s S$ and $\operatorname{T}_t T$ as indicated.
If you go through the calculations, you will find that
$$\begin{align}
\nabla R &= 0
\\ \operatorname{Ric}\left((X_s,0),(X_s,0)\right) &=
   \left\langle(X_s,0),(X_s,0)\right\rangle_p
   \neq 0 &\text{for }& X_s\neq 0
\\ \operatorname{Ric}\left((0,X_t),(0,X_t)\right) &=
   0 \neq \left\langle(0,X_t),(0,X_t)\right\rangle_p
   &\text{for }& X_t\neq 0
\end{align}$$
which shows that $M$ is locally symmetric, but not an Einstein manifold.
Now you might suspect that the statement of your question was only a near miss,
and you might want to additionally require some sort of
irreducibility of the Riemannian manifold $M$.
You may also want to require that $M$ be complete and simply connected
to get a global symmetric space from the local one.
Then look up e.g. the following:


*

*Chapter XI (symmetric spaces), theorem 8.6 + corollary 8.7 from: Shoshichi Kobayashi and Katsumi Nomizu: Foundations of differential geometry, Vol.2. Wiley 1969

*Section 2, corollary 3 from these Lecture notes on symmetric spaces by J.-H. Eschenburg
These concretize what irreducibility of $M$ shall mean and ultimately
establish an Einstein manifold.
