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# 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.

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

## Problem

Let be a polynomial with integer coefficients,such that, are odd integers.Show that

(a) does not have any even integer roots.

(b) 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 has no integer roots,then we are done.

Let us assume has an integer root .

Then we can write, where, is any function of .

Now ,putting in ,we get and, now as and ,are consecutive integers and cannot be both odd,

which means that cannot be both odd,given whatever are.

So,we can conclude that there exists no integer solution of .

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

## Problem

Let be a polynomial with integer coefficients,such that, are odd integers.Show that

(a) does not have any even integer roots.

(b) 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 has no integer roots,then we are done.

Let us assume has an integer root .

Then we can write, where, is any function of .

Now ,putting in ,we get and, now as and ,are consecutive integers and cannot be both odd,

which means that cannot be both odd,given whatever are.

So,we can conclude that there exists no integer solution of .

Hence,we are done.

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