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Explore the Back-StoryThis 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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

** 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.**

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.

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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

** 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.**

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.

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