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
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.
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 $P(1)P(8)P(9) = 1988$.
