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# Proving Irreducibles

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.

Note by Sharky Kesa
2 years, 6 months ago

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Let $f(x) = (x-a_1)(x-a_2)(x-a_3)..........(x-a_n) - 1$ 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. · 2 years, 6 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 $$(x+2)(x+2)-1=(x+1)(x+3)$$. · 2 years, 6 months ago

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