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

Related

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

Related

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

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

HALL OF FAME BLOG OFFLINE CLASS

CHEENTA ON DEMAND BOSE OLYMPIAD

Cheenta Academy Private Limited |

support@cheenta.com

Menu

ISI CMI Program