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Registered User
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Mar 31 |
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irreducible polynomials with arithmetic progression coefficients I mean $2n+1$,the leading coefficient instead of $n+1$ in the above |
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Mar 31 |
comment |
irreducible polynomials with arithmetic progression coefficients Since the coefficients is increasing, all the roots of $g(x)$ have absolute value less than $1$ so that $g$ is irreducible if $n+1$ is prime and in general $g$ has at most $m$ irreducible factors if $n+1$ has $m$ prime factors (counting multiplities). |
