Test of Mathematics Solution Subjective 175 - Integer Roots

This is a Test of Mathematics Solution Subjective 175 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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Problem

Let $\text{P(x)}=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_{1}x+a_{0}$ be a polynomial with integer coefficients,such that,$\text{P(0) and P(1)}$ are odd integers.Show that

(a)$\text{P(x)}$ does not have any even integer roots.

(b)$\text{P(x)}$ does not have any odd integer roots.

Solution

Given the two statements (a) and (b) above it is clear that if we can prove that $\text{P(x)}$ has no integer roots,then we are done.

Let us assume $\text{P(x)}$ has an integer root $\text{ a}$.

Then we can write,

$$ \text{P(x)=(x-a)Q(x)}\dots(1)$$

where,$\text{Q}$is any function of $\text{ x}$.

Now ,putting $\text{x=0,1}$ in $\text{ 1}$,we get

$$\text{P(0)=(-a)Q(0)}\dots(2)$$

and,

$$\text{P(1)=(1-a)Q(1)}\dots(3)$$

now as $\text{(1-a)}$ and $\ \text{(-a)}$,are consecutive integers $ \text{(2)}$ and $\text{(3)}$

cannot be both odd,

which means that $\text{P(0),P(1)}$ cannot be both odd,given whatever $\text{Q(0) and Q(1)}$ are.

So,it is a contradiction!!.

So,we can conclude that there exists no integer solution of $\text{P(x)}$.

Hence,we are done.

Problem

Let $\text{P(x)}=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_{1}x+a_{0}$ be a polynomial with integer coefficients,such that,$\text{P(0) and P(1)}$ are odd integers.Show that

(a)$\text{P(x)}$ does not have any even integer roots.

(b)$\text{P(x)}$ does not have any odd integer roots.