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Is X8+1X^8+1 reducible in R[x]\mathbb R[x]?

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Prove or disprove: x2016+1∈R[x]x^{2016} + 1 \in \mathbb R[x] is irreducible.

Can I just say that b2−4ac<0b^2-4ac<0 so is irreducible? ================= ================= 3 Answers 3 ================= It's not a quadratic polynomial, so you can't use its discriminant (at least in elementary number theory). It has no real root, but it splits into a product of linear factors in C[x]\mathbf C[x], and the corresponding complex roots are pairwise conjugate since this is a polynomial with real coefficients. Furthermore, these roots have modulus 11. Grouping the conjugate factors (x−ζ)(x−ˉζ)=x2−2(Reζ)x+1,(x-\zeta)(x-\bar\zeta)=x^2-2(\operatorname{Re}\zeta) x+1, you obtain a decomposition into the product of 10081008 irreducible quadratic factors. Note: It is a theorem that the irreducible polynomials in R[x]\mathbf R[x] are : 1) linear polynomial; 2) quadratic polynomials with a negative discriminant. Hint: Show that x2016+1x^{2016} + 1 is divisible by x32+1x^{32} + 1 (no matter what the base ring RR actually is). x2016+1x^{2016} + 1 is reducible, since it's the sum of two cubes.