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Showing tan((p+1) theta) has numerator divisible by p

Source: Brazilian Undergrad Mathematics Olympiad 2022 P6

November 23, 2022
number theoryprime numbers

Problem Statement

Let p≡3 (mod 4)p \equiv 3 \,(\textrm{mod}\, 4) be a prime and θ\theta some angle such that tan⁡(θ)\tan(\theta) is rational. Prove that tan⁡((p+1)θ)\tan((p+1)\theta) is a rational number with numerator divisible by pp, that is, tan⁡((p+1)θ)=uv\tan((p+1)\theta) = \frac{u}{v} with u,v∈Z,v>0,mdc(u,v)=1u, v \in \mathbb{Z}, v >0, \textrm{mdc}(u, v) = 1 and u≡0 (mod p)u \equiv 0 \,(\textrm{mod}\,p) .
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