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I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

Update: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but more elegant form,

$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?

I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

Update 1: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but more elegant form,

$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$

Or more generally (for $\mu$ an integer),

$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde} = 0$$

where $\mu^5$ is the constant term of the quintic and the Lehmer quintic the special case $\mu=1$.

Update 2: I've already tested the Hashimoto septic which fortunately has a seventh power $\mu^7$ as its constant term and the analogous relation,

$$\frac{\mu}{a}-\frac{\mu^{2}}{ab}+\frac{\mu^{3}}{abc}-\frac{\mu^{4}}{abcd}+\frac{\mu^{5}}{abcde}-\frac{\mu^{6}}{abcdef}+\frac{\mu^{7}}{abcdefg} = 0$$

and among the $(p-1)! = 720$ permutation of its roots, at least 9 works.

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?

I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

Update: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but much simpler form,

$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?

I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

Update: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but more elegant form,

$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?

I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?

I. Comparison

It doesn't seem to be well-known that the generic cubic (prominent in this MO post) for $C_3 = A_3$,

$$x^3-nx^2+(n-3)x+1 = 0$$

has the nice property that its roots $a,b,c$, if in correct order, obey,

$$(a^2b)^{1/3}+(b^2c)^{1/3}+(c^2a)^{1/3} = 0$$

(I only noticed this after I asked an MO question about the similar-looking Klein quartic $a^3b+b^3c+c^3a=0.$)

Since the generic cubic is intimately connected to the roots of unity for prime $p\equiv 1\,\text{mod}\; 6$, (the case $n=1$ yields $1^{1/7}$), naturally I got curious about its big sister the Emma Lehmer quintic which is for $p\equiv 1\,\text{mod}\; 10$, namely,

$$x^5 + n^2x^4 - (2n^3 + 6n^2 + 10n + 10)x^3 + (n^4 + 5n^3 + 11n^2 + 15n + 5)x^2 + (n^3 + 4n^2 + 10n + 10)x +1=0$$

It turns out its roots $x_k$ may have a similar property.

II. Question:

Analogous to the generic cubic, is it true that the Emma Lehmer quintic obeys, $$(a^4b^3c^2d)^{1/5} + (b^4c^3d^2e)^{1/5} + (c^4d^3e^2a)^{1/5} + (d^4e^3a^2b)^{1/5} + (e^4a^3b^2c)^{1/5} = 0$$ for the correct ordering of its roots $a,b,c,d,e$?

Update: Thanks to Peter Taylor in the comments, and using the fact that $abcde = -1$, we can get rid of the fifth roots and get the equivalent but much simpler form,

$$\frac1{a}-\frac1{ab}+\frac1{abc}-\frac1{abcd}+\frac1{abcde} = 0$$

III. Example

Let $n=-1$. Then we get the quintic for $p=11$ and its roots,

$$x^5 + x^4 - 4x^3 - 3x^2 + 3x + 1 = 0$$

$$a,b,c,d,e = 2\cos\frac{2\pi k}{11}$$

with $k = 1, 4, 5, 2, 3$ as the correct order. One can then verify it obeys the relation in the question.

P.S. For $p=5$, there are $(p-1)! = 24$ permutations of its roots. It is easy for a computer to find the correct order for any $n$ that I tested. But does it hold true for ALL $n$?