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