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irreducibility of binomials with unity
coefficients∗
pahio†
2013-03-21 19:06:25
Let n be a positive integer. We consider the possible factorization of the
binomial xn +1.
• If n has no odd prime factors, then the binomial xn + 1 is irreducible.
Thus, x+1, x2 +1, x4 +1, x8 +1 and so on are irreducible polynomials (i.e.
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irreducible in the field Q of their coefficients). N.B., only
√ x+1 and x√+1 are
irreducible in the field R; e.g. one has x4+1 = (x2−x 2+1)(x2+x 2+1).
• If n is an odd number, then xn +1 is always divisible by x+1:
xn + 1 = (x + 1)(xn−1 − xn−2 + xn−3 − + · · · − x + 1)
(1)
This formula is usable when n is an odd prime number, e.g.
x5 + 1 = (x + 1)(x4 − x3 + x2 − x + 1).
• When n is not a prime number but has an odd prime factor p, say n = mp,
then we write xn +1 = (xm )p +1 and apply the idea of (1); for example:
x12 + 1 = (x4 )3 + 1 = (x4 + 1)[(x4 )2 − x4 + 1] = (x4 + 1)(x8 − x4 + 1)
There are similar results for the binomial xn + y n , and the formula corresponding to (1) is
xn + y n = (x + y)(xn−1 − xn−2 y + xn−3 y 2 − + · · · − xy n−2 + y n ),
(2)
which may be verified by performing the multiplication on the right hand side.
∗ hIrreducibilityOfBinomialsWithUnityCoefficientsi
created: h2013-03-21i by: hpahioi
version: h36982i Privacy setting: h1i hResulti h12D05i h13F15i
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