Polynomials and Arithmetic: If $p(x_i) = 7$ for four integers, then $p(z)\ne14$

Question

Consider the polynomial
$$p(x) = a_0 + a_1x + a_2x^2 + · · · + a_nx^n$$ where $a_0, a_1, . . . , a_n ∈ \Bbb Z$. Show that if $p(x_i) = 7$ for 4 distinct integers $x_0, x_1, x_2, x_3$, then $p(z) \neq 14$ for any $z ∈ \Bbb Z$.

How do I start this question?

Answer 1

Let $q(x)=p(x)-7$. Now the problem becomes:

Show that if $q(x_i)=0$ for $4$ distinct integers $x_0,x_1,x_2,x_3,$ then $q(z) \ne 7$ for any $z \in \mathbb Z$

But if $q(x_i) = 0$ for $i=0,\ldots,3,$ then

$$q(x)=(x-x_0)(x-x_1)(x-x_2)(x-x_3)r(x)$$

for some integer polynomial $r(x)$. Suppose $q(z) = 7$ for some $z \in \mathbb Z$. Can you see the contradiction?