Find that number

We only need to know that $P(a_1)=P(a_2)=P(a_3)=P(a_4)=10$ for four distinct integers $a_1,a_2,a_3,a_4$ . We can find a monic polynomial $F(X)$ with integer coefficients such that $P(X) - 10 \; = \; F(X)(X-a_1)(X-a_2)(X-a_3)(X-a_4)$ and hence the integer $b$ , if it existed, would be such that $F(b)(b-a_1)(b-a_2)(b-a_3)(b-a_4) \; = \; 981-10 = 971$ Since $971$ is prime, this means that $b-a_1,b-a_2,b-a_3,b-a_4$ are integers that divide $971$ , so are all equal to one of $1,-1,971,-971$ . Since their product divides $971$ , we deduce that at least three of $a_1-b,a_2-b,a_3-b,a_4-b$ must be equal to either $1$ or $-1$ , and hence that at least two of $a_1-b,a_2-b,a_3-b,a_4-b$ are equal to eacn other (being both equal to either $1$ or $-1$ ). This means that at least two of $a_1,a_2,a_3,a_4$ are equal to each other, which is a contradiction.

Thus no such integer $b$ exists, making the answer $\boxed{0}$ .

It is kind of obvious that it should be evaluated in 1 and the coefficients should be 1. It means that the polynomial is x^10 + x^9 + ... + x. So it si not posible to be equal yo 981.