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I assume you want $q$ and $r$ to be odd primes. Also, note that I will be using the notation that $\zeta_m$ means an arbitrary primitive $m$-th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality.

Lemma: For any odd $m>1$ and any $\zeta_m$, the number $\zeta_m+1$ is a unit.

Proof: Let $r$ be such that $m | 2^r-1$. We'll abbreviate $\zeta_m$ to $\zeta$.

Then $\zeta^{2^r} = \zeta$ so $$1 = \left( \frac{\zeta^{2}-1}{\zeta -1} \right) \left( \frac{\zeta^{4}-1}{\zeta^{2} -1} \right) \cdots \left( \frac{\zeta^{2^r}-1}{\zeta^{2^{r-1}} -1} \right)=$$ $$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r-1}} +1 \right),$$ exhibiting an explicit inverse for $\zeta+1$.

Note that, for $m$ odd, the negation of a primitive

Let $2m$-th root of unity is \eta$be a primitive$m$-th 2qr$ root of unity. So the negation of Then your proposed unit is $\zeta_q+\zeta_{-q} + \zeta_{r} + \zeta_{-r}$. There is a primitive $qr$-th root of unity, $\eta$, such that $\zeta_q = \eta^r$ and $\zeta_r = \eta^q$. So the claim is that $$\eta^r + \eta^{-r} \eta^{r}+\eta^{-r} + \eta^q + \eta^{-q}$$ is a unit. This eta^{-q}$and factors as $$\eta^r \left(1 + \eta^{q-r} +\eta^{-q-r} + \eta^{-2r} \right) = \eta^r \left( 1+ \eta^{q-r} \right) \left( 1+\eta^{-q-r} \right).$$ Using that (1+\eta^{q-r})(1+\eta^{-q-r}).$$Since q and r are primes, we have GCD(q-r, qr) = GCD(q+r,qr)=1odd and relatively prime, so \eta^{q-r} and \eta^{q+r} are primitive qr-th roots of unity , and we are done by the lemma. Note: I don't think I used that q and r were odd primes, only that they were odd, relatively prime, and >1. 1 I assume you want q and r to be odd primes. Also, note that I will be using the notation that \zeta_m means an arbitrary primitive m-th root of unity (but the same one every time it appears in an equation), and will be proving the statement in that generality. Lemma: For any odd m>1 and any \zeta_m, the number \zeta_m+1 is a unit. Proof: Let r be such that m | 2^r-1. We'll abbreviate \zeta_m to \zeta. Then \zeta^{2^r} = \zeta so $$1 = \left( \frac{\zeta^{2}-1}{\zeta -1} \right) \left( \frac{\zeta^{4}-1}{\zeta^{2} -1} \right) \cdots \left( \frac{\zeta^{2^r}-1}{\zeta^{2^{r-1}} -1} \right)=$$ $$\left( \zeta+1 \right) \left( \zeta^{2} + 1 \right) \cdots \left( \zeta^{2^{r-1}} +1 \right),$$ exhibiting an explicit inverse for \zeta+1. Note that, for m odd, the negation of a primitive 2m-th root of unity is a primitive m-th root of unity. So the negation of your proposed unit is \zeta_q+\zeta_{-q} + \zeta_{r} + \zeta_{-r}. There is a primitive qr-th root of unity, \eta, such that \zeta_q = \eta^r and \zeta_r = \eta^q. So the claim is that$$\eta^r + \eta^{-r} + \eta^q + \eta^{-q}$$is a unit. This factors as $$\eta^r \left(1 + \eta^{q-r} +\eta^{-q-r} + \eta^{-2r} \right) = \eta^r \left( 1+ \eta^{q-r} \right) \left( 1+\eta^{-q-r} \right).$$ Using that$q$and$r$are primes, we have$GCD(q-r, qr) = GCD(q+r,qr)=1$, so$\eta^{q-r}$and$\eta^{q+r}$are primitive$qr$-th roots of unity, and we are done by the lemma. Note: I don't think I used that$q$and$r$were odd primes, only that they were odd, relatively prime, and$>1\$.