Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 18H / b / Solution

A primitive tenth root $\xi$ has ${\xi }^5=-1$ and $\xi \ne -1$, so division of $X^5+1$ by $X+1$ givesIt is irreducible over $\mathbb{Q}$: replacing $X$ by $-X$ gives ${\Phi }_5(X)$, and then replacing $X$ by $X+1$ gives $X^4+5X^3+10X^2+10X+5$. Any nontrivial monic integer factorization would reduce modulo five to factors that are powers of $X$, forcing both constant terms divisible by five and their product divisible by $25$, contrary to the constant term five. Rational factorization reduces to integer factorization by Gauss lemma: products of primitive integer polynomials are primitive, since reduction modulo any prime preserves a nonzero product. ThusLet $t=\xi +{\xi }^{-1}$. Dividing ${\Phi }_{10}(\xi )=0$ by ${\xi }^2$ gives $t^2-t-1=0$. Hence $(2t-1{)}^2=5$, so $\boxed{\sqrt{5}\in \mathbb{Q}(\xi )}$. Either sign of this square root suffices; for $\xi =e^{\pi i/5}$ the positive one is $2t-1$.