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I believe the only change needed to handle all cases (even and odd) is to replace the exponent (q**d-1)/2 with (q**d-1)/(the smallest prime factor of q**d-1), although there are more efficient techniques.
[[User:MeanStandev|MeanStandev]] ([[User talk:MeanStandev|talk]]) 20:55, 27 April 2018 (UTC)
== factorization of polynomial in a ring ''Z''/15''Z'' ==
In ''Z''/15''Z'', (''x''+1) × (''x''+2) = ''x''<sup>2</sup> + 3''x'' + 2, and (''x''+7) × (''x''+11) is also ''x''<sup>2</sup> + 3''x'' + 2, and all of the polynomials ''x''+1, ''x''+2, ''x''+7 and ''x''+11 are irreducible (since all of they have degree 1), thus ''x''<sup>2</sup> + 3''x'' + 2 have two factorizations to irreducible polynomials in ''Z''/15''Z''. Besides, in ''Z''/15''Z'', ''x''<sup>2</sup> + 2 is irreducible, since −2 is not a quadratic residue mod 15.
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