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Lemma: If $p=2r+1$, then $U:≔\prod_{k=1}^r (1-z^{2k-1})$ solves $N(U)=±I$ provided $r$ is odd.

Indeed, the argument in the footnote above shows that $U$ is the product over primitive roots $ζ$ of $1$ of degree $4p$ in the first quadrant, hence $N(U)$ is $\Pi$.

 Theorem: the equation $N(u) = ±I$ has a solution which is a cyclotomic unit iff there is a odd power $2^{2k+1}≕2P$ of $2$ which is $±1 \pmod p$.

Since the phase of $N(1-ζ)$ is $±π/2$, this follows from:

Lemma: If $ξ ≔ N(u)$ with a cyclotomic unit $u$ of $K$ is a root of $1$, then $ξ^4=1$. Moreover, the set of such $u$ (up to roots of $1$) is generated²⁾ by $S(ζ) :≔ (1-ζ)(1-ζ^2)(1-ζ^4)(1-ζ^8)\cdots(1-ζ^P)$, where $ζ$ is a primitive roots of $1$ of degree $4p$ and $P$ is a power of $2$ such that $2P \equiv ±1 \pmod p$.

²⁾ There is only one (non-trivial) relation. If $r$ and $s$ are invertible residues mod $4p$, say that $z^r∼z^s$ if $$\{±r, ±2r, ±4r, ±8r, \ldots, ±Pr\} = \{±s, ±2s, ±4s, ±8s, \ldots, ±Ps\}.$$ Then $S(z^r)$ and $S(z^s)$ coincide up to roots of unity. Moreover, the only relation between $S(ζ)$ with $ζ:=z^r$ and $r$ running through these equivalence classes is that their product is a root of unity.

Above we saw that $N(1-ζ)$ is (up to a root of $1$) a $p$-cyclotomic unit. Indeed, $N(1-ζ)=1/ζ-ζ=(1+(-ζ^2))/ζ$, and $-ζ^2$ is a primitive root of $1$ of degree $p$. If $-ζ^2=z^{4t}$ for some $t$, then $1-ζ^2=(1-(-ζ^2)^2)/(1-(-ζ^2))=ε_{2t}/ε_t$, where $ε_r:≔(1-ρ^r)/(1-ρ)$ are $p$-cyclotomic units defined for $r\not\equiv 0 \pmod p$, with $\rho^p=1$. Recall that the units $ε_r$ and $ε_{-r}$ coincide modulo roots of $1$, moreover the only other relation modulo roots of $1$ is $ε_1=1$.

Essentially, the index $r$ lives in the cyclic group $(ℤ/p)^\star/\{±1\}$. Hence the condition of the lemma holds if the multiplicity of $t$ “cancels” the multiplicity of $2t$ , except for the case $2t\equiv 1 \pmod p$. However, it is easy to see that then this holds also for $2t\equiv 1 \pmod p$ ,  which implies the lemma.

Lemma: If $p=2r+1$, then $U:≔\prod_{k=1}^r (1-z^{2k-1})$ solves $N(U)=±I$ provided $r$ is odd. Update: moreover, if $Λ$ is the (multiplicative) lattice generated by $(1-ζ)/(1-(-ζ^{-1}))$ with primitive roots ζ in the first quadrant. then $N(Λ)=1$, and any solution modulo this lattice is $U^s$ with an odd $s$.

Indeed, the argument in the footnote above shows that $U$ is the product over primitive roots $ζ$ of $1$ of degree $4p$ in the first quadrant, hence $N(U)$ is $\Pi$. (Update: this also describes all the solutions to $N(f)=\text{unit}$ for any $p$. In particular, if $4|p-1$, then $s∉ℤ$, hence there is no solution in cyclotomic units.)

Update: here was a completely wrong “theorem” about the case $4|p-1$. The arguments above show only this:

If the index of cyclotomic units inside units of $K$ is odd, there is no solution. (While vol.1 of Lang’s Cyclotomic fields contains some info about the 2-part of this index, I’m not fluent enough to use this info. I do not even know how to find it in PARI/gp.)

Th: the equation $N(u) = ±I$ has a solution which is a cyclotomic unit iff there is a odd power $2^{2k+1}≕2P$ of $2$ which is $±1 \pmod p$.

²⁾ There is only one (non-trivial) relation. If $r$ and $s$ are invertible residues mod $4p$, say that $z^r∼z^s$ if $$\{±r, ±2r, ±4r, ±8r, \ldots, ±Pr\} = \{±s, ±2s, ±4s, ±8s, \ldots, ±Ps\}.$$ Then $S(z^r)$ and $S(z^s)$ coincide up to roots of unity. Moreover, the only relation between $S(ζ)$ with $ζ:=zʳ$ and $r$ running through these equivalence classes is that their product is a root of unity.

Theorem: the equation $N(u) = ±I$ has a solution which is a cyclotomic unit iff there is a odd power $2^{2k+1}≕2P$ of $2$ which is $±1 \pmod p$.

²⁾ There is only one (non-trivial) relation. If $r$ and $s$ are invertible residues mod $4p$, say that $z^r∼z^s$ if $$\{±r, ±2r, ±4r, ±8r, \ldots, ±Pr\} = \{±s, ±2s, ±4s, ±8s, \ldots, ±Ps\}.$$ Then $S(z^r)$ and $S(z^s)$ coincide up to roots of unity. Moreover, the only relation between $S(ζ)$ with $ζ:=z^r$ and $r$ running through these equivalence classes is that their product is a root of unity.

This sketch of a half-answer is based on and is developing the ideas of Max’ answer. He works with ℚ[I,x] with I²=-1 modulo the cyclotomic polynomial Φₚ(x). Writing x as z⁴ and I is zᵖ, this is identified with the 4p-cyclotomic field K, i.e., ℚ[z] modulo Φ₄ₚ(z). In particular, Max’ F is identified with an element f∈K.

Since his F˟(x) = xᵖ⁻²F(1/x), his x²⁻ᵖF˟ is identified with σ·f for a suitable involution of K. One can see that σz=z²ᵖ⁻¹=-z⁻¹ (so σI=I, σx=1/x). Denote by N₂ the norm N(g)≔g·σg of K over the fixed points K₂ of σ. Conclusion: Max’ equation is equivalent to N(f) = -2(z²+1/z²)² (with integer f).

Since N(I)=N(z)=-1, N(1+I)=2I, and N(1+x)=(z²+1/z²)², it is enough to solve N(g)=±I. Recall that since 4p is not a power of a prime, cyclotomic units are generated (multiplicatively) by the units 1-ζ (for primitive roots ζ of 1 of degree 4p), and roots of 1 in K. Modulo roots of 1, they form a lattice spanned by such 1-ζ with Im ζ>0 and the only relation¹⁾ ∏(1-ζ)=I⁽ᵖ⁻¹⁾ˊ². (Their importance is in the fact that they have finite index in units of K.)

¹⁾ Indeed, if Π is this product, then |Π|² = Φ₄ₚ(1) = 1 (since Φ₄ₚ(x)(x²+1)(x²ᵖ-1)=x⁴ᵖ-1), and combining 1-ζ and 1+1/ζ together gives N(1-ζ)=1/ζ-ζ with argument -π/2. Hence argument of Π is (p-1)π/4.

Lemma: If p=2r+1, then U≔∏ₖ₌₁ʳ (1-z²ᵏ⁻¹) solves N(U)=±I provided r is odd.

Indeed, the argument in the footnote above shows that U is the product over primitive roots ζ of 1 of degree 4p in the first quadrant, hence N(U) is Π.

Th: the equation N(u) = ±I has a solution which is a cyclotomic unit iff there is a odd power 2²ᵏ⁺¹≕2P of 2 which is ±1 mod p.

Since the phase of N(1-ζ) is ±π/2, this follows from:

Lemma: If ξ ≔ N(u) with a cyclotomic unit u of K is a root of 1, then ξ⁴=1. Moreover, the set of such u is (up to roots of 1) is generated²⁾ by S(ζ) ≔ (1-ζ)(1-ζ²)(1-ζ⁴)(1-ζ⁸)…(1-ζᴾ) with ζ a primitive roots of 1 of degree 4p and P a power of 2 such that 2P ≡ ±1 mod p.

²⁾ There is only one (non-trivial) relation. If r and s are invertible residues mod 4p, say that zʳ∼zˢ if {±r, ±2r, ±4r, ±8r, …, ±Pr} = {±s, ±2s, ±4s, ±8s, …, ±Ps}. Then S(zʳ) and S(zˢ) coincide up to roots of unity. Moreover, the only relation between S(ζ) with ζ=zʳ and r running through these equivalence classes is that their product is a root of unity.

Above we saw that N(1-ζ) is (up to a root of 1) a p-cyclotomic unit. Indeed, N(1-ζ)=1/ζ-ζ=(1+(-ζ²))/ζ, and -ζ² is a primitive root of 1 of degree p. If -ζ²=z⁴ᵗ, then 1-ζ²=(1-(-ζ²)²)/(1-(-ζ²))=ε₂ₜ/εₜ. Here εᵣ≔(1-ρʳ)/(1-ρ) are p-cyclotomic units defined for r≢0 mod p, with ρᵖ=1. Recall that the units εᵣ and ε₋ᵣ coincide modulo roots of 1, moreover the only other relation modulo roots of 1 is ε₁=1.

Essentially, the index r lives in the cyclic group (ℤ/p)˟/{±1}. Hence the condition of the lemma holds if the multiplicity of t “cancels” the multiplicity of 2t — except for the case 2t≡1 mod p. However, it is easy to see that then this holds also for 2t≡1 mod p — which implies the lemma.

This sketch of a half-answer is based on and is developing the ideas of Max’ answer. He works with $\mathbb Q[I,x]$ with $I^2=-1$ modulo the cyclotomic polynomial $\Phi_p(x)$. Writing $x$ as $z^4$ and $I$ is $z^p$, this is identified with the $4p$-cyclotomic field $K$, i.e., $\mathbb Q[z]$ modulo $\Phi_{4p}(z)$. In particular, Max’ $F$ is identified with an element $f\in K$.

Since his $F^\star(x) = x^{p-2}F(1/x)$, his $x^{2-p}F^\star$ is identified with $\sigma\cdot f$ for a suitable involution $\sigma$ of $K$. One can see that $σz=z^{2p-1}=-z^{-1}$ (so $σI=I$, $σx=1/x$). Denote by $N$ the norm $N(g)≔g·σg$ of $K$ over the fixed points $K_2$ of $σ$. Conclusion: Max’ equation is equivalent to $N(f) = -2(z^2+1/z^2)^2$ (with integer $f$).

Since $N(I)=N(z)=-1$, $N(1+I)=2I$, and $N(1+x)=(z^2+1/z^2)^2$, it is enough to solve $N(g)=\pm I$. Recall that since $4p$ is not a power of a prime, cyclotomic units are generated (multiplicatively) by the units $1-ζ$ (for primitive roots $ζ$ of $1$ of degree $4p$), and roots of $1$ in $K$. Modulo roots of $1$, they form a lattice spanned by such $1-ζ$ with $\Im ζ>0$ and the only relation¹⁾ $\prod (1-ζ)=I^{(p-1)/2}$. (Their importance is in the fact that they have finite index in units of $K$.)

¹⁾ Indeed, if $\Pi$ is this product, then $|\Pi|^2 = \Phi_{4p}(1) = 1$ (since $\Phi_{4p}(x)(x^2+1)(x^{2p}-1)=x^{4p}-1$), and combining $1-ζ$ and $1+1/ζ$ together gives $N(1-ζ)=1/ζ-ζ$ with argument $-π/2$. Hence argument of $\Pi$ is $(p-1)π/4$.

Lemma: If $p=2r+1$, then $U:≔\prod_{k=1}^r (1-z^{2k-1})$ solves $N(U)=±I$ provided $r$ is odd.

Indeed, the argument in the footnote above shows that $U$ is the product over primitive roots $ζ$ of $1$ of degree $4p$ in the first quadrant, hence $N(U)$ is $\Pi$.

Th: the equation $N(u) = ±I$ has a solution which is a cyclotomic unit iff there is a odd power $2^{2k+1}≕2P$ of $2$ which is $±1 \pmod p$.

Since the phase of $N(1-ζ)$ is $±π/2$, this follows from:

Lemma: If $ξ ≔ N(u)$ with a cyclotomic unit $u$ of $K$ is a root of $1$, then $ξ^4=1$. Moreover, the set of such $u$ (up to roots of $1$) is generated²⁾ by $S(ζ) :≔ (1-ζ)(1-ζ^2)(1-ζ^4)(1-ζ^8)\cdots(1-ζ^P)$, where $ζ$ is a primitive roots of $1$ of degree $4p$ and $P$ is a power of $2$ such that $2P \equiv ±1 \pmod p$.

²⁾ There is only one (non-trivial) relation. If $r$ and $s$ are invertible residues mod $4p$, say that $z^r∼z^s$ if $$\{±r, ±2r, ±4r, ±8r, \ldots, ±Pr\} = \{±s, ±2s, ±4s, ±8s, \ldots, ±Ps\}.$$ Then $S(z^r)$ and $S(z^s)$ coincide up to roots of unity. Moreover, the only relation between $S(ζ)$ with $ζ:=zʳ$ and $r$ running through these equivalence classes is that their product is a root of unity.

Above we saw that $N(1-ζ)$ is (up to a root of $1$) a $p$-cyclotomic unit. Indeed, $N(1-ζ)=1/ζ-ζ=(1+(-ζ^2))/ζ$, and $-ζ^2$ is a primitive root of $1$ of degree $p$. If $-ζ^2=z^{4t}$ for some $t$, then $1-ζ^2=(1-(-ζ^2)^2)/(1-(-ζ^2))=ε_{2t}/ε_t$, where $ε_r:≔(1-ρ^r)/(1-ρ)$ are $p$-cyclotomic units defined for $r\not\equiv 0 \pmod p$, with $\rho^p=1$. Recall that the units $ε_r$ and $ε_{-r}$ coincide modulo roots of $1$, moreover the only other relation modulo roots of $1$ is $ε_1=1$.

Essentially, the index $r$ lives in the cyclic group $(ℤ/p)^\star/\{±1\}$. Hence the condition of the lemma holds if the multiplicity of $t$ “cancels” the multiplicity of $2t$ , except for the case $2t\equiv 1 \pmod p$. However, it is easy to see that then this holds also for $2t\equiv 1 \pmod p$ ,  which implies the lemma.