Unstable

Algebra Level 3

Which of the expressions below is factorizable?

Q 1 ( p ) = [ p ( 2 m + 1 ) n ] [ p ( 2 m 1 ) n ] . . . [ p n ] [ p + n ] . . . [ p + ( 2 m 1 ) n ] [ p + ( 2 m + 1 ) n ] 1 Q_1(p) = [p - (2m + 1)n][p - (2m - 1)n]...[p - n][p + n]...[p + (2m - 1)n][p + (2m + 1)n] - 1

Q 2 ( p ) = [ p ( 2 m + 1 ) n ] 2 [ p ( 2 m 1 ) n ] 2 . . . [ p n ] 2 [ p + n ] 2 . . . [ p + ( 2 m 1 ) n ] 2 [ p + ( 2 m + 1 ) n ] 2 + 1 Q_2(p) = [p - (2m + 1)n]^2[p - (2m - 1)n]^2...[p - n]^2[p + n]^2...[p + (2m - 1)n]^2[p + (2m + 1)n]^2 + 1


Note: 0 m n p Z 0 \ne m \ne n \ne p \in \mathbb Z .


This is part of the series: " It's easy, believe me! "

None of the fuctions is factorizable. Q 2 ( p ) Q_2(p) All of the fuctions is factorizable. Q 1 ( p ) Q_1(p)

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1 solution

Thành Đạt Lê
Sep 30, 2017

This is not actually a solution, but rather a generalization, that:

( x a 1 ) ( x a 2 ) . . . ( x a n 1 ) ( x a n ) 1 (x - a_1)(x - a_2)...(x - a_{n - 1})(x - a_n) - 1 and ( x a 1 ) 2 ( x a 2 ) 2 . . . ( x a n 1 ) 2 ( x a n ) 2 + 1 (x - a_1)^2(x - a_2)^2...(x - a_{n - 1})^2(x - a_n)^2 + 1 are both unfactorizable in Z [ x ] \mathbb Z[x] , n N n \in \mathbb N^* and a 1 a 2 a n 1 a n Z a_1 \ne a_2 \ne \cdots \ne a_{n - 1} \ne a_{n} \in \mathbb Z .

Q 1 ( p ) Q_1(p) and Q 2 ( p ) Q_2(p) are just a small part in this qeneralization, making a problem more specific sometimes makes the problem harder, right?

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