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Görre Mörre (talk | contribs) m →Square-free factorization: A small improvement to a comment in the psuedocode. |
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* On any field extension of '''F'''<sub>2</sub>, ''P'' = (''x''+1)<sup>4</sup>.
*On every other finite field, at least one of −1, 2 and −2 is a square<ref>Why? None of −1, 2 and −2 is a square mod 15</ref>, because the product of two non-squares is a square and so we have
#If <math>-1=a^2,</math> then <math>P=(x^2+a)(x^2-a).</math>
#If <math>2=b^2,</math> then <math>P=(x^2+bx+1)(x^2-bx+1).</math>
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