Show there are $b, c \in \mathbb{R}$ such that $f(x)= {a\over 2}x^2 + bx + c$ 
Given $f:I\rightarrow \mathbb{R}$ and $f''(x) = a.\forall x\in I$.
  Show there are $b, c \in \mathbb{R}$ such that $f(x)= {a\over 2}x^2  + bx + c$.

If we define $g(x) = ax + b$, then $g'(x) = f''(x) = a$. Therefore, $f'(x)$ and $g(x)$ are equal (maybe differ by a constant). Hence, $f'(x)$ has to have the form: $f'(x) = ax + b$
I've been told to use Lagrange's Mean Value Thm but I'm not sure why the above alone isn't sufficient.
P.S
I wrote "Therefore, $f'(x)$ and $g(x)$ are equal (maybe differ by a constant)". How should I write it in a more professional/mathematical way?
 A: I  would just say that given $f''(x) = a$, and integrating with respect to $x$ twice will give you the desired result.  $\int f''(x) dx= \int a dx = ax + b$, where $b$ is a constant. Then, $\int f'(x) dx= \int ax + b dx = {(ax^2)\over2} + bx + c$, where $c$ is a constant.  and this is the form you wanted. 
A: A partial answer: 
For $f'(x)$: Consider any two $x_1$ and $x_2$, by LMVT, $\frac{f'(x_1)-f'(x_2)}{x_1-x_2} = a $, which is the equation for a straight line, with slope $a$, and so $f'(x) = ax + b$. I am not sure if the next extension is as simple. I'll try to complete the rest and edit this reply.
[Edit]:
At $x_3$ and $x_4$, for $f(x)$, by LMVT:  $\frac{f(x_3)-f(x_4)}{x_3-x_4} = a\frac{x_3+x_4}{2}+b $, let $x_4 = 0$ and $x_3=x$. Then, $\frac{f(x)-f(0)}{x} = a\frac{x}{2}+b$ which gives, $f(x) = \frac{ax^2}{2}+bx + c$ where, $c=f(0)$
I think even for $f'(x)$ computation, setting $x_2=0$ and $x_1=x$ would be better, than directly using straight line equation.
$$
\frac{f'(x_1)-f'(x_2)}{x_1-x_2} = a, with x_1=x, x_2=0
$$ 
becomes
$$
\frac{f'(x)-f'(0)}{x} = a \implies f'(x) = ax+b, b=f'(0)
$$
PS: Thanks a lot for an interesting approach, that most of us wouldn't have thought of.
