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Yaakov Baruch
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A family of counterexamples is defined as follows:

$r=2$,

$p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$,

$a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)} \qquad b=\dfrac{2\tan(\pi d_1/p_1)\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)}$,

(with $a=1, \;b=2\tan(\pi d_2/p_2)\;$ if $\;p_1=2$)

$g(X)=(X-(a+i b)^2)(X-(a-ib)^2)=X^2-2(a^2-b^2)X+(a^2+b^2)^2$,

$p=\text{lcm}(p_1, p_2), \quad q=2$,

$\prod_{j=0}^{p} f(\omega^j X)=(X^{p}-(a-1+ib)^{p})^2(X^{p}-(-a-1+ib)^{p})^2$.

Here is a picture of the four roots of $f$ in the complex plane:  The four roots of <span class=$f$ " />


Here is a picture of the four roots of $f$ in the complex plane: enter image description here

When $p_1\!=\!p_2$, $\;g$ is a square and the formulas simplify to the original answer:

$r\!=\!2, \; p\!\ge\!3, \; b\!=\!\tan(\frac{\pi}{p}), \; g(X)\!=\!X\!+b^2, \; q\!=\!2,\; \prod\limits_j f(\omega^j X)\!=\!\big(X^p\!-(-1\!+ib)^p\big)^2$.


When $\;a=\pm b\;$ the example simplifies to: $\; r=4, \; g=X+4a^4, \quad$ but the only such cases I know of are for $\;(d_1, p_1, d_2, p_2) \!=\! (3, 8, 1, 8)\;$ and $\;\prod f(\omega^j X)=(X^8-(\frac{1+i}{\sqrt{2}}-1)^8)^2(X^8-(\frac{-1+i}{\sqrt{2}}-1)^8)^2$,$\;\;$ or $(d_1, p_1, d_2, p_2) \!=\! (1, 6, 1, 12)\;$ and $\;\prod f(\omega^j X)=(X^{12}-(\frac{-1+i}{1+\sqrt{3}}-1)^{12})^2(X^{12}-(\frac{1+i}{1+\sqrt{3}}-1)^{12})^2\;$.


Ignoring trivial derivations (i.e. replacing $g$ with products of the "primitive" ones above), I'd be very surprised if any other examples esisted - but a proof of that looks to me extremely hard.

A family of counterexamples is defined as follows:

$r=2$,

$p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$,

$a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)} \qquad b=\dfrac{2\tan(\pi d_1/p_1)\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)}$,

(with $a=1, \;b=2\tan(\pi d_2/p_2)\;$ if $\;p_1=2$)

$g(X)=(X-(a+i b)^2)(X-(a-ib)^2)=X^2-2(a^2-b^2)X+(a^2+b^2)^2$,

$p=\text{lcm}(p_1, p_2), \quad q=2$,

$\prod_{j=0}^{p} f(\omega^j X)=(X^{p}-(a-1+ib)^{p})^2(X^{p}-(-a-1+ib)^{p})^2$.

Here is a picture of the four roots of $f$ in the complex plane:  The four roots of <span class=$f$ " />


When $p_1\!=\!p_2$, $\;g$ is a square and the formulas simplify to the original answer:

$r\!=\!2, \; p\!\ge\!3, \; b\!=\!\tan(\frac{\pi}{p}), \; g(X)\!=\!X\!+b^2, \; q\!=\!2,\; \prod\limits_j f(\omega^j X)\!=\!\big(X^p\!-(-1\!+ib)^p\big)^2$.


When $\;a=\pm b\;$ the example simplifies to: $\; r=4, \; g=X+4a^4, \quad$ but the only such cases I know of are for $\;(d_1, p_1, d_2, p_2) \!=\! (3, 8, 1, 8)\;$ and $\;\prod f(\omega^j X)=(X^8-(\frac{1+i}{\sqrt{2}}-1)^8)^2(X^8-(\frac{-1+i}{\sqrt{2}}-1)^8)^2$,$\;\;$ or $(d_1, p_1, d_2, p_2) \!=\! (1, 6, 1, 12)\;$ and $\;\prod f(\omega^j X)=(X^{12}-(\frac{-1+i}{1+\sqrt{3}}-1)^{12})^2(X^{12}-(\frac{1+i}{1+\sqrt{3}}-1)^{12})^2\;$.


Ignoring trivial derivations (i.e. replacing $g$ with products of the "primitive" ones above), I'd be very surprised if any other examples esisted - but a proof of that looks to me extremely hard.

A family of counterexamples is defined as follows:

$r=2$,

$p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$,

$a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)} \qquad b=\dfrac{2\tan(\pi d_1/p_1)\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)}$,

(with $a=1, \;b=2\tan(\pi d_2/p_2)\;$ if $\;p_1=2$)

$g(X)=(X-(a+i b)^2)(X-(a-ib)^2)=X^2-2(a^2-b^2)X+(a^2+b^2)^2$,

$p=\text{lcm}(p_1, p_2), \quad q=2$,

$\prod_{j=0}^{p} f(\omega^j X)=(X^{p}-(a-1+ib)^{p})^2(X^{p}-(-a-1+ib)^{p})^2$.

Here is a picture of the four roots of $f$ in the complex plane: enter image description here

When $p_1\!=\!p_2$, $\;g$ is a square and the formulas simplify to the original answer:

$r\!=\!2, \; p\!\ge\!3, \; b\!=\!\tan(\frac{\pi}{p}), \; g(X)\!=\!X\!+b^2, \; q\!=\!2,\; \prod\limits_j f(\omega^j X)\!=\!\big(X^p\!-(-1\!+ib)^p\big)^2$.


When $\;a=\pm b\;$ the example simplifies to: $\; r=4, \; g=X+4a^4, \quad$ but the only such cases I know of are for $\;(d_1, p_1, d_2, p_2) \!=\! (3, 8, 1, 8)\;$ and $\;\prod f(\omega^j X)=(X^8-(\frac{1+i}{\sqrt{2}}-1)^8)^2(X^8-(\frac{-1+i}{\sqrt{2}}-1)^8)^2$,$\;\;$ or $(d_1, p_1, d_2, p_2) \!=\! (1, 6, 1, 12)\;$ and $\;\prod f(\omega^j X)=(X^{12}-(\frac{-1+i}{1+\sqrt{3}}-1)^{12})^2(X^{12}-(\frac{1+i}{1+\sqrt{3}}-1)^{12})^2\;$.


Ignoring trivial derivations (i.e. replacing $g$ with products of the "primitive" ones above), I'd be very surprised if any other examples esisted - but a proof of that looks to me extremely hard.

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Yaakov Baruch
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Let A family of counterexamples is defined as follows:

$r=2$,

$p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$,

$a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)} \qquad b=\dfrac{2\tan(\pi d_1/p_1)\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)}$,

(with $g(X)=X+3$$a=1, \;b=2\tan(\pi d_2/p_2)\;$ if $\;p_1=2$)

$g(X)=(X-(a+i b)^2)(X-(a-ib)^2)=X^2-2(a^2-b^2)X+(a^2+b^2)^2$,

$p=\text{lcm}(p_1, p_2), \quad q=2$,

$\prod_{j=0}^{p} f(\omega^j X)=(X^{p}-(a-1+ib)^{p})^2(X^{p}-(-a-1+ib)^{p})^2$.

Here is a picture of the four roots of $p=3$;$f$ in the complex plane:  The four roots of <span class=$f$ " $f(X)=X^2+2X+4$/>


When $p_1\!=\!p_2$, $\;g$ is not a square, and the formulas simplify to the original answer:

$r\!=\!2, \; p\!\ge\!3, \; b\!=\!\tan(\frac{\pi}{p}), \; g(X)\!=\!X\!+b^2, \; q\!=\!2,\; \prod\limits_j f(\omega^j X)\!=\!\big(X^p\!-(-1\!+ib)^p\big)^2$.


When $\;a=\pm b\;$ the example simplifies to: $\; r=4, \; g=X+4a^4, \quad$ but the only such cases I know of are for $f(X)f(\omega X)f(\omega^2 X)=(X^3-8)^2$ is$\;(d_1, p_1, d_2, p_2) \!=\! (3, 8, 1, 8)\;$ and $\;\prod f(\omega^j X)=(X^8-(\frac{1+i}{\sqrt{2}}-1)^8)^2(X^8-(\frac{-1+i}{\sqrt{2}}-1)^8)^2$,$\;\;$ or $(d_1, p_1, d_2, p_2) \!=\! (1, 6, 1, 12)\;$ and $\;\prod f(\omega^j X)=(X^{12}-(\frac{-1+i}{1+\sqrt{3}}-1)^{12})^2(X^{12}-(\frac{1+i}{1+\sqrt{3}}-1)^{12})^2\;$.


Ignoring trivial derivations (i.e. replacing $g$ with products of the "primitive" ones above), I'd be very surprised if any other examples esisted - but a proof of that looks to me extremely hard.

Let $r=2$, $g(X)=X+3$, $p=3$; $f(X)=X^2+2X+4$ is not a square, but $f(X)f(\omega X)f(\omega^2 X)=(X^3-8)^2$ is.

A family of counterexamples is defined as follows:

$r=2$,

$p_1\ge 2, \quad p_2\ge 3, \quad d_1 \perp p_1, \quad d_2 \perp p_2$,

$a=\dfrac{\tan(\pi d_1/p_1)-\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)} \qquad b=\dfrac{2\tan(\pi d_1/p_1)\tan(\pi d_2/p_2)}{\tan(\pi d_1/p_1)+\tan(\pi d_2/p_2)}$,

(with $a=1, \;b=2\tan(\pi d_2/p_2)\;$ if $\;p_1=2$)

$g(X)=(X-(a+i b)^2)(X-(a-ib)^2)=X^2-2(a^2-b^2)X+(a^2+b^2)^2$,

$p=\text{lcm}(p_1, p_2), \quad q=2$,

$\prod_{j=0}^{p} f(\omega^j X)=(X^{p}-(a-1+ib)^{p})^2(X^{p}-(-a-1+ib)^{p})^2$.

Here is a picture of the four roots of $f$ in the complex plane:  The four roots of <span class=$f$ " />


When $p_1\!=\!p_2$, $\;g$ is a square and the formulas simplify to the original answer:

$r\!=\!2, \; p\!\ge\!3, \; b\!=\!\tan(\frac{\pi}{p}), \; g(X)\!=\!X\!+b^2, \; q\!=\!2,\; \prod\limits_j f(\omega^j X)\!=\!\big(X^p\!-(-1\!+ib)^p\big)^2$.


When $\;a=\pm b\;$ the example simplifies to: $\; r=4, \; g=X+4a^4, \quad$ but the only such cases I know of are for $\;(d_1, p_1, d_2, p_2) \!=\! (3, 8, 1, 8)\;$ and $\;\prod f(\omega^j X)=(X^8-(\frac{1+i}{\sqrt{2}}-1)^8)^2(X^8-(\frac{-1+i}{\sqrt{2}}-1)^8)^2$,$\;\;$ or $(d_1, p_1, d_2, p_2) \!=\! (1, 6, 1, 12)\;$ and $\;\prod f(\omega^j X)=(X^{12}-(\frac{-1+i}{1+\sqrt{3}}-1)^{12})^2(X^{12}-(\frac{1+i}{1+\sqrt{3}}-1)^{12})^2\;$.


Ignoring trivial derivations (i.e. replacing $g$ with products of the "primitive" ones above), I'd be very surprised if any other examples esisted - but a proof of that looks to me extremely hard.

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Yaakov Baruch
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Let $r=2$, $g(X)=X+3$, $p=3$; $f(X)=X^2+2X+4$ is not a square, but $f(X)f(\omega X)f(\omega^2 X)=(X^3-8)^2$ is.