Does the polynomial $P(x)$ have integer zeros?

The following is a homework question:

Let $P(x)$ be a polynomial with integer coefficients and $P(x_1)=P(x_2)=P(x_3)=P(x_4)=P(x_5)=P(x_6)=P(x_7)=7$ where $x_i$ are distinct integers. Determine if $P(x)$ has integer zeros.

I've never done questions like this before. I started with this:

If $\deg(P) = 7$,

$$P(x)=\alpha(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$$

where $\alpha$ is some integer.

However, the question doesn't state that the polynomial must be of the seventh degree. Even then, I don't see how I can determine if $P(x)$ has integer zeros without knowing all the $x_i$.

Can someone please help me? Thanks.

Edit:
Is this a valid solution?

$$P(x)=Q(x)(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$$

If $P(n)=0$,
$$Q(n)(n-x_1)(n-x_2)(n-x_3)(n-x_4)(n-x_5)(n-x_6)(n-x_7)=-7$$

where $Q(x)$ is a polynomial of integer coefficients (therefore $Q(n)$ is an integer)

And since all the terms on the LHS are integers and the $x_i$ are distinct integers, it follows that some of the factors on the LHS $\ne$ {$\pm 1, \pm 7$}. And $7$ is a prime number, therefore $n$ cannot be a zero.

Answer

Hint:

$P(x)= Q(x) (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)(x-x_6)(x-x_7)+7$

Idea : You can't get $-7$ by multiplying $7$ distinct integers..