Prove $f(b)\equiv f(c)$(mod $m$) Is this sufficient in proving the following statement? Also is there a more efficient way of doing so? Thanks in advance.

Prove: If $f(x) = a_nx^n +\dots  +a_1x+a_0$ is a polynomial with integer
  coecients and $b \equiv c $(mod $m)$ then $f(b)\equiv f(c)$(mod $m$).

By Induction of degree $f$

Base Case: 

Suppose $f$ is of degree $0$.
Then for some constant $d$, $f(x)=d$,   $\forall x$.
Thus if $b \equiv c$, then $f(b)=d=f(c)\Rightarrow f(b)\equiv f(c)$

Induction hypothesis:

Suppose this works for all polynomials of degree $k$.

Inductive Step:

Let $f$ be a polynomial of degree $k+1$
$f(x)=a_{k+1}x^{k+1}+a_kx^k+\dots +a_1x+a_0$
$x(a_{k+1}x^{k}+a_kx^{k-1}+\dots +a_1)+a_0$
For some $a_0,\dots,a_{k+1}$, where $a_{k+1}$ is non-zero.
Let $g(x)=a_{k+1}x^{k}+a_kx^{k-1}+\dots +a_1$
Then $g(x)$ is of degree $k$, since $a_{k+1}$ does not equal $0$ and $f(x)=xg(x)+a_0$
By Induction hypothesis:
$b\equiv c\Rightarrow g(b)\equiv g(c)$ 
$\Rightarrow bg(b)\equiv bg(c)$ 
$\Rightarrow bg(b)\equiv cg(c)$ 
$\Rightarrow bg(b)+a_0\equiv cg(c)+a_0$ 
$\Rightarrow f(b)\equiv f(c)$ 
$\therefore f(b)\equiv f(c)$(mod$m$) $\forall$ polynomials $f$.
 A: HINT:
Group all the summands in the difference $f(b) - f(c)$ and then use the fact that $b-c$ is always a factor of $b^n - c^n; \forall n \in \mathbb{N}$, hence all of the "grouped" summands are divisible by $m$.
In addition your proof seems to be alright.
A: Write $c=b+tm$. Then the binomial theorem implies that $c^k=b^k+t_km$.
A: The following are well known properties of clock arithmetic:
$(1). b=c\mod m\Rightarrow b^n=c^n\mod m$
$(2). b=c\mod m\Rightarrow ab=ac\mod m$
$(3). b_1=c_1\mod m\text{ and }b_2=c_2\mod m\Rightarrow b_1+b_2=c_1+c_2\mod m.$
Using those properties it is inmediate to prove what you want.
Edit: As it seems to be some confussion I sketch the proof here: Take $k$ an integer such that $0\leq k\leq n$. By property (1) it holds that $b^k=c^k\mod m$. By property (2) it holds that $a_kb^k=a_kc^k\mod m$. Now, property (3) is obviously true for 3, 4,... or any finite number of summands. So sum the equations for $k=0,1,...,n$ and you get $f(b)=f(c)\mod m$.
Notice that the only "induction argument" used here is that (3) can be generalized to an arbitrary (finite) number of summands.
