$f(x)$ be a polynomial with integer coefficients and $f(0) = 1989$ and $f(1) = 9891$, then no. of polynomial

Question

Let $f(x)$ be a polynomial with integer coefficients and $f(0) = 1989$ and $f(1) = 9891$. Then prove that $f(x)$ has no integer roots.

$\bf{My\; Try::}$ Let $f(x) = a_{0}x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+...........+a_{n}\;,$ where $a_{0},a_{1},.......,a_{n}\in \mathbb{Z}$

Using Division Algorithm.

$f(x) = x\cdot q_{1}(x)+1989$ and $f(x) = (x-1)\cdot q_{2}(x)+9891\;,$ Where $q_{1}(x)$ and $q_{2}(x)$

are polynomial with integer Coefficients.

So $\displaystyle x\cdot \left(q_{1}(x)-q_{2}(x)\right)+q_{2}(x)-7902 = 0\Rightarrow x\cdot \left(q_{1}(x)-q_{2}(x)\right) = q_{2}(x)-7902$.

Put $x=0\;,$ we get $q_{2}(0) = 7902$.

Put $x=1\;,$ we get $q_{1}(1)-q_{2}(1) = q_{2}(1)-7902\Rightarrow 2q_{2}(1)-q_{1}(1) = 7902$

Now How can i prove it, Thanks

Answer 1

Hint: If $m\equiv n\pmod q$, then $f(m)\equiv f(n)\pmod q$.

Try $q=2$.

A more direct proof would look at:

$$g(x)=f(x)-1989-7902x$$

Which must be divisible $x(x-1)$. So $g(x)$ is always even for integer $x$, and hence $f(x)=g(x)+7902x+1989$ is always odd form integer $x$.