polynomials $P(x)$ with integer coefficients such that $P(1)\cdot P(9)\cdot P(8) = 1988$ Calculation of polynomials $P(x)$ with integer coefficients such that $P(1)\cdot P(9)\cdot P(8) = 1988$
$\bf{My\; Try}::$ Given $P(1)\cdot P(9)\cdot P(8) = 1988 = 2^2\cdot 7 \cdot 71$
and Let $P(x) = a_{0}+a_{1}x+a_{2}x^2+...........+a_{n}x^n$
Now I did not understand how can i solve after that
Help required
Thanks
 A: Hint: If $a$ and $b$ are integers, then $a-b | P(a) - P(b)$.
Unfortunately, after this, there is some brute work that needs to be done, so I'm not sure how helpful of a hint it is.

In light of bashing needed, you are better off just applying the technique of Lagrange Interpolation Formula directly, to the finitely many cases of factorization.
A: Start with Calvin's hint: $a - b\,|\,P(a)-P(b)\;\forall a,b\in \mathbb{Z}$, we have
$$7\,|\,P(8) - P(1)\quad\text{ and }\quad 8\,|\,P(9) - P(1)$$
There are two consequences


*

*$7\not|\,P(1)$ and $7 \not|\,P(8)$   
Otherwise, $7\,|\,P(8)-P(1) \implies 7^2 | P(1)P(8) \implies 7^2|1988$, a contradiction!

*$4\not|\,P(1)$ and $4 \not|\,P(9)$
Otherwise, $8\,|\,P(9)-P(1) \implies 2^4 | P(1)P(9) \implies 2^4 |1988$, another contradiction!
So in terms of where the factors go, there are only following 24 possibilites:
$$( P(1),P(8),P(9) ) = \text{ one of }\left\{\begin{array}{lll}
(\pm 2 \cdot 71,& \pm 1,  & \varepsilon 14 ),\\
(\pm 2,         & \pm 71, & \varepsilon 14),\\
(\pm 2,         & \pm 1,  & \varepsilon 14\cdot 71)\\
(\pm 1 \cdot 71,& \pm 4,  & \varepsilon 7),\\
(\pm 1,         & \pm 4\cdot 71,& \varepsilon 7)\\
(\pm 1,         & \pm 4,   & \varepsilon 7\cdot 71)
\end{array}\right.$$
where $\varepsilon = \text{sign}(P(1)P(8))$.
Notice $71 \equiv 1 \pmod 7$. If one look at everything modulus 7, we get
$$(P(1),P(8)) \equiv (\pm 2, \pm 1) \text{ or } (\pm 1, \pm 4) \pmod 7
\quad\implies\quad P(1) \not\equiv P(8)\pmod 7$$
This contradict with the requirement $7\,|\,P(8)-P(1)$ and hence there is 
no integer polynomial $P(x)$ with $P(1)P(8)P(9) = 1988$.
