Let ${\mathrm{a}}_0,{\mathrm{a}}_1,{\mathrm{a}}_{2,}\dots \dots ,{\mathrm{a}}_{\mathrm{n}}$ be integers and $\mathrm{P}\left(\mathrm{x}\right)={\mathrm{a}}_0{\mathrm{x}}^{\mathrm{n}}+{\mathrm{a}}_1{\mathrm{x}}^{\mathrm{n}-1}+\dots .+{\mathrm{a}}_{\mathrm{n}}$ such that P(1),P(0) are odd then the number of integral solutions of equation P(x) = 0 is

Hints: P(0) = an is an odd integer; $\mathrm{P}\left(1\right)={\mathrm{a}}_0+{\mathrm{a}}_1+\dots {\mathrm{a}}_{\mathrm{n}}$ is an odd integer. 
Let $\alpha$ be an even integer then
$\mathrm{P}\left(\mathrm{\alpha }\right)=\left({\mathrm{a}}_0{\mathrm{\alpha }}^{\mathrm{n}}+\dots ..+{\mathrm{a}}_{\mathrm{n}-1}\mathrm{\alpha }\right)+{\mathrm{a}}_{\mathrm{n}}$ is odd
$\Rightarrow \mathrm{P}\left(\mathrm{\alpha }\right)\ne 0$ 0 Let $\beta$ be an odd integer then
${\text{P(β)=a}}_0\left({\mathrm{\beta }}^{\mathrm{n}}-1\right)+{\mathrm{a}}_1\left({\mathrm{\beta }}^{\mathrm{n}-1}-1\right)+\dots +{\mathrm{a}}_{\mathrm{n}-1}(\mathrm{\beta }-1)+{\mathrm{a}}_0+{\mathrm{a}}_1+\dots {\mathrm{a}}_{\mathrm{n}}$
is also odd then $\mathrm{P}\left(\mathrm{\beta }\right)\ne 0$.
Hence there is no integral solution