Are you thinking, it's infinite.

$f(x)=a_{0}+a_{1}x+a_{2}x^{2}+.......$

• x=7 : $11=a_{0}+7a_{1}+49a_{2}+......$
• x=11 : $13=a_{0}+11a_{1}+121a_{2}+.......$

Subtracting first equation from second, $2=4a_{1}+72a_{2}+......$

Here we can see that no integral set of values of $a_{n}$ exists satisfying this equation!

f(y)-f(x) must be divisible by (y-x)

If all the coefficients are integers

But there is a contradiction when y=11 x=7

This will use proof by contradiction.

Assume that such a polynomial exist.

$g(x) := f(x+7)-11$ is also a polynomial with integral coefficients.

We have: $\begin{cases} g(0) &=& 0 \\\ g(4) &=& 2 \end{cases}$

Composing the above polynomial with the map $\Bbb Z \to \Bbb Z / 4 \Bbb Z$ gives us the contradiction $0 \equiv 2$ .

(11-7) divides f (11)-f (7)....what else needed??