Show $a - b \mid f(a) - f(b)$ for $f(x)$ an integer coefficient polynomial [Polynomial Factor Theorem] I have seen this lemma elsewhere. 
Suppose $A$ is a domain, and $f \in A[X]$. Prove that
$$a - b \mid f(a) - f(b)$$

I need to prove this. 
$$f(a) - f(b) \equiv 0 \pmod{a-b}$$ basically.
Let, $a - b = c$
$$f(a) - f(b)/(a-b) = f'(\xi)$$ for Some $\xi \in (a, b)$.
But I dont see it showing divisibility 
 A: Let $A$ be a domain with field of fractions $K$ and let
$$
f(X)=a_0+a_1X+a_2X^2+\cdots+a_nX^n
$$
be a polynomial in $A[X]$. Then for all $a\neq b\in A$,
$$
\begin{eqnarray}
f(b)-f(a)&=&(a_0+a_1b+\cdots+a_nb^n)-(a_0+a_1a+\cdots+a_na^n)\\
 &=& (b-a)\left(a_1+a_2\frac{b^2-a^2}{b-a}+\cdots+a_n\frac{b^n-a^n}{b-a}\right)
\end{eqnarray}
$$
is certainly valid in $K$. But now note that both factors actually belong to $A$ since $b-a$ divides $b^k-a^k$ for all $k\geq0$.
A: This follows by specializing the Polynomial Factor Theorem (below), namely if $\,f\in A[x]\,$ then
$$\begin{align} 
x\!-\!b\ \ \ \mid\  &\ f(x)-f(b)\ \ \ {\rm in}\ \ A[x]\\[.2em]
\Rightarrow\ \ g(x)\, (x\!-\!b) =&\ f(x)-f(b)\,\ \ {\rm for\ some}\ \ g\in A[x]\\[.2em] 
\Rightarrow\ \ g(a)\, (a\!-\!b)\, =&\ f(a)-f(b)\,\ \ {\rm by\ eval\ at}\ \ x=a\\[.2em] 
\Rightarrow\qquad\ \ \ \ \, a\!-\!b\ \ \ \,\mid\  &\, f(a)-f(b)\ \ \ {\rm in}\ \ A\ \end{align}\ \ \ $$
Polynomial Factor Theorem $\ $ If $A$ is a commutative ring, $\,b\in A\,$ and $\,f(x)\in A[x]\,$ then
$$\ x\!-\!b\,\mid\, f(x)-f(b)\ \ {\rm in}\ \ A[x]\quad $$
Proof $\ \ {\rm mod}\,\ x\!-\!b\!:\,\ \color{#c00}{x\equiv b}\,\Rightarrow\, f(\color{#c00}{x})\equiv f(\color{#c00}b)\ $ by the Polynomial Congruence Rule. (the linked proof of the congruence rule is given in $\,\Bbb Z\,$ but it is in fact valid in any commutative ring, since the proofs  use only commutative ring axioms).
Or  use the Polynomial Division Algorithm to write $\,f(x)-f(b) = (x\!-\!b)g(x) + r\,$ then conclude $\,r = 0\,$ by evaluation at $\, x = b\,$ (recall eval is a ring hom when the coef ring $A$ is commutative).
Or, directly, use linearity to reduce to the monomial case $\,f = x^k\,$ then use the high-school formula for $\,(x^n-b^n)/(x-b)\,$ to infer it is $\in A[x],\,$ i.e. explicitly
$$\dfrac{f(x)-f(b)}{x-b} = \sum \,f_i\, \dfrac{x^k - b^k}{x-b} = \sum \,f_i\,(x^{k-1}\! +\! bx^{k-2}\!+\cdots+\! b^{k-2}x\!+\!b^{k-1})\in A[x] $$
Remark $ $ The universal Polynomial Factor Theorem $\ x\!-\!y\,\mid\, f(x)-f(y)\in \Bbb Z[x,y]\,$ is the special case  $\, A=\Bbb Z[y]\,$ and $\,b = y.\,$ This allows us to deduce all "number" instances as specializations of "function" (polynomial) cases. It is a simple prototypical example of deducing number divisibility as a special case of  polynomial divisibility. Here is more on universality of polynomial identities.
We have the following equivalence between the above methods.
Theorem $\ $ TFAE for a polynomial $\,f\in R[x],\,$ and $\,a\in R\,$ a commutative ring.
$(0)\ \ \ f = (x\!-\!a)q + r\ $ for some $\,q\in R[x],\ r\in R\ \ \ $ [Monic Linear Division Algorithm]
$(1)\ \ \ f\bmod x\!-\!a = f(a)\ \ \ \ \ \ \ $ [Remainder Theorem]
$(2)\ \ \ f(a) = 0\,\Rightarrow\, x\!-\!a\mid f\ \ \ \ $ [Factor Theorem]
Proof $\ (0\Rightarrow 1)\ \ \ f = (x\!-\!a)q + r\,\overset{\large x\,=\,a}\Longrightarrow\, r=f(a)\,\Rightarrow\,f\bmod x\!-\!a = r = f(a) $
$(1\Rightarrow 2)\ \ \ f\bmod x\!-\!a = f(a) = 0\,\Rightarrow\, x\!-\!a\mid f$
$(2\Rightarrow 0)\ \ \ g := f-f(a)\,$ has $\,g(a) = 0\ $ so $\ g = f-f(a) = (x\!-\!a)q$
A: I'll assume that with $f$ you mean a polynomial with integer coefficients.
By setting $k=a-b$, it is sufficient to prove that $f(b+k)-f(b)$ is divisible by $k$. Now, consider the polynomial in two variables
$$
g(x,y)=f(x+y)-f(x)
$$
Then $g(x,0)=0$, which means that $g(x,y)$ is divisible by $y$ in the polynomial ring $\mathbb{Z}[x,y]$; thus $g(x,y)=yh(x,y)$ for some $h\in\mathbb{Z}[x,y]$.
Therefore
$$
f(b+k)-f(b)=g(b,k)=kh(b,k)
$$
is divisible by $k$ as required.
A: The universal property of the polynomial ring $A[X]$ asserts that for every $A$-algebra $C$ and $c \in C$, there exists a unique ring homomorphism $\text{ev}_c : A[X] \to C$ which sends $X$ to $c$.
If $f \in A[X]$ is a polynomial, we commonly denote $\text{ev}_c(f) = f(c)$.
Let $\pi : A \to A/(a-b)$ be the projection.
Take $B = A$ with $c = a$ and $B = A$ with $c = B$. This gives us two ring homomorphisms $\pi \circ \text{ev}_a, \pi \circ \text{ev}_b$, which both send $X$ to $\pi(a) = \pi(b)$. From the uniqueness in the universal property, it follows that
$$\pi \circ \text{ev}_a = \text{ev}_{\pi(a)} = \pi \circ \text{ev}_b$$
That is, $\pi(f(a)) = \pi(f(b))$ for all $f \in A[X]$, so that $f(a) \equiv f(b) \pmod{a-b}$.
