For a  polynomial $P$ for which $P(x+5) - P(x) = 2 ,\forall x$. What is the least possible value of $P(4) - P(2)$? There are  infinite number of polynomials $P$ for which $P(x+5) - P(x) = 2,\forall x \in \mathbb{R}$. 
How could we determine the least possible value of $P(4) - P(2)$?
 A: Notice that $P(5)=2+P(0)$, $P(10)=4+P(0)$ and, more generally, $$(\star)\qquad\qquad P(5k)=2k+P(0)$$ for all $k\in\mathbb N$.
Now, a little calculus shows that

the degree $d$ of a polynomial $f$ is the unique element of $\mathbb N_0$ such that $\lim\limits_{k\to\infty}\frac{f(5k)}{(5k)^d}$ is a non-zero (finite!) real number, and moreover the value of that limit is the coefficient of $x^d$ in $f$.

Using $(\star)$ we see easily that $\lim\limits_{k\to\infty}\frac{P(5k)}{5k}=\frac{2}{5}$: it follows that $P$ is a polynomial of degree $1$ of the form $P(x)=\frac{2}{5}x+b$ for some $b\in\mathbb R$.
But then we can completely compute $P(4)-P(2)$ to find its value $\frac{4}{5}$.
A: Let $\mathrm{P}_n(\mathbb{R})$ be the vector space of polynomials $f$ with real coefficients and $\deg f\le n$, i.e. of the form
$$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,\quad a_i\in\mathbb{R}.$$
Now the delta operator $\Delta_h:f(x)\mapsto f(x+h)-f(x)$ for $h\ne0$ is a projection from $\mathrm{P}_n(\mathbb{R})$ down to the subspace $\mathrm{P}_{n-1}(\mathbb{R})$, $n\ge1$: the leading terms in $f(x+h)$ and $f(x)$ cancel each other out. Thus
$$\Delta_h f\in \mathrm{P}_n(\mathbb{R})\iff f\in \mathrm{P}_{n+1}(\mathbb{R}).$$
In particular, this tells us that if $\Delta_5 f=2$ is a constant polynomial, $f$ is linear $f(x)=ax+b$. From this it is easy to see that $\Delta_5 f\,(0)= a\cdot5=2$ implies $a=2/5$, whence $f(4)-f(2)$ is $\frac{2}{5}(4-2)=\frac{4}{5}$.
A: Hint: for a polynomial $f(x)$ over a field $K$ of characteristic $ =0,\:$ and $\:0\ne k\in K$
$$ \deg f \le 1\:  \iff\: f(x+k) - f(x) \in K\: \iff\: \dfrac{f(x+k) - f(x)}k\: =\: f'(0)\:=\:\dfrac{f(4)-f(2)}2 $$
A: The condition $P(x+5)-P(x)=2$ for all $x$ can be rewritten as "$(x,y)$ is on the graph of $y=P(x)$ if and only if $(x+5,y+2)$ is."  This implies that if $(x,y)$ is on the graph of $y=P(x)$, so are $(x+5k,y+2k)$ for all $k\in\mathbb{Z}$.  All of those points lie on a line of the form $y=\frac{2}{5}x+b$ for some $b\in\mathbb{R}$, so that line intersects the graph of $y=P(x)$ infinitely many times, or equivalently, $P(x)=\frac{2}{5}x+b$ has infinitely many solutions.
The only way for that to happen with a polynomial $P$ is if $P(x)=\frac{2}{5}x+b$ for all $x\in\mathbb{R}$ (two distinct polynomials can only be equal at finitely many points—if two polynomials are equal at infinitely many points, they must be identically equal).  Thus, $$P(4)-P(2)=\left(\frac{2}{5}\cdot 4+b\right)-\left(\frac{2}{5}\cdot 2+b\right)=\frac{4}{5}.$$
A: If you know Calculus:
For each $x$ there exists some $c_x \in (x,x+5)$ so that 
$$\frac{P(x+5)-P(x)}{5}=P'(c_x) \,.$$
Thus $P'(c_x)=\frac{2}{5}$.
Thus,the polynomial $P'(x)$ takes the value $\frac{2}{5}$ infinitely many times (at least once in any interval of length 5), and thus 
$$P'(x)=\frac{2}{5} \,.$$
Thus 
$$P(x)=\frac{2}{5}x+b \,,$$
for some $b$.
