Prove that the polynomial

\((x-a_1)(x-a_2)...(x-a_n)-1\),

where \(a_1, a_2, ..., a_n\) are distinct integers, cannot be written as the product of two non-constant polynomials with integer coefficients, i.e., it is irreducible.

Prove that the polynomial

\((x-a_1)(x-a_2)...(x-a_n)-1\),

where \(a_1, a_2, ..., a_n\) are distinct integers, cannot be written as the product of two non-constant polynomials with integer coefficients, i.e., it is irreducible.

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TopNewestLet \[f(x) = (x-a_1)(x-a_2)(x-a_3)..........(x-a_n) - 1\] For sake of contradiction let us consider \(f(x)\) can be written as the product of two polynomials with integer co-efficients,ie,\(f(x) = p(x)q(x)\). For all i = 1,...,n we have \[f(a_i) = p(a_i)q(a_i) = -1\]. Since \(p(x)\) and \(q(x)\) are polynomials with integers co-efficients. we must have \(p(a_i) = 1\) and \(q(a_i) = -1\) OR \(p(a_i) = -1\) and \(q(a_i) = 1\)

In either case we have \[p(a_i)+q(a_i) = 0\]. Hence The polynomial \(p(x) + q(x)\) must have n distinct roots. But the degrees of both \(p(x)\) and \(q(x)\) are less than n and hence \(p(x)+ q(x)\) cannot have n roots. Hence Contradiction. – Eddie The Head · 2 years, 9 months ago

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An interesting note here: Dropping the "distinct integer" restriction on the problem invalidates the claim. One simple counterexample is \((x+2)(x+2)-1=(x+1)(x+3)\). – Daniel Liu · 2 years, 9 months ago

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