How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

Test of Mathematics Solution Subjective 175 - Integer Roots

Test of Mathematics at the 10+2 Level

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




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.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.