 How Cheenta works to ensure student success?
Explore the Back-Story

# Test of Mathematics Solution Subjective 43-Integer Root  This is a Test of Mathematics Solution Subjective 43 (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

Show that the equation $x^3 + 7x - 14(n^2 +1) = 0$ has no integral root for any integer n.

## Solution:

We note that $14(n^2 +1) - 7x = x^3$ implies $x^3$ is divisible by 7. This implies x is divisible by 7 (as 7 is a prime number). Suppose x= 7x'. Hence we can rewrite the given equation as:

$7^3 x'^3 + 7 \times 7 x' - 14 (n^2 +1 ) = 0$.

Cancelling out a 7 we have $7^2 {x'}^3 + 7{x'} = 2(n^2 +1)$. Since 7 divides left hand side, it must also divide the right hand side. Since 7 cannot divide 2, it must divide $n^2 + 1$ as 7 and 2 are coprime.  Note that 7 cannot divide $n^2 +1$ as square of a number always gives remainder 0, 1, 4, 2 when divided by 7 and never 6. But if $n^2 + 1$ is divisible by 7 then $n^2$ must give remainder 6 when divided by 7.  Hence contradiction.

Necessary Lemma: square of a number always gives remainder 0, 1, 4, 2 when divided by 7

$n \equiv 0, \pm 1 , \pm 2 , \pm 3 \mod 7\Rightarrow n^2 \equiv 0, 1, 4, 9 (=2) \mod 7$

Key Ideas: Modular Arithmetic This is a Test of Mathematics Solution Subjective 43 (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

Show that the equation $x^3 + 7x - 14(n^2 +1) = 0$ has no integral root for any integer n.

## Solution:

We note that $14(n^2 +1) - 7x = x^3$ implies $x^3$ is divisible by 7. This implies x is divisible by 7 (as 7 is a prime number). Suppose x= 7x'. Hence we can rewrite the given equation as:

$7^3 x'^3 + 7 \times 7 x' - 14 (n^2 +1 ) = 0$.

Cancelling out a 7 we have $7^2 {x'}^3 + 7{x'} = 2(n^2 +1)$. Since 7 divides left hand side, it must also divide the right hand side. Since 7 cannot divide 2, it must divide $n^2 + 1$ as 7 and 2 are coprime.  Note that 7 cannot divide $n^2 +1$ as square of a number always gives remainder 0, 1, 4, 2 when divided by 7 and never 6. But if $n^2 + 1$ is divisible by 7 then $n^2$ must give remainder 6 when divided by 7.  Hence contradiction.

Necessary Lemma: square of a number always gives remainder 0, 1, 4, 2 when divided by 7

$n \equiv 0, \pm 1 , \pm 2 , \pm 3 \mod 7\Rightarrow n^2 \equiv 0, 1, 4, 9 (=2) \mod 7$

Key Ideas: Modular Arithmetic

### Knowledge Partner  