Re: Digest for competition-corner-participant-discussion@googlegroups.com - 1 update in 1 topic

[Elkies, Noam]

Tom <t...@bellefleurbooks.com> wrote:

> I had thought that you could show that s is in the field as well as c
> and then make a relatively simple heptagon to demonstrate the
> conjecture. As it turns out, by going down this path, you do not find
> that s is in the field. Rather, as it turns out, you find that (7^0.5)s
> is in the field.

Yes, I in effect noted this already: s^2 is 1-c^2, which is 1/4 times
what I called 4-d^2, and the second of my "further remarks" was
the observation that (4-d^2) (d^2-d-2)^2 = 7, which means that
2s or -2s is sqrt(7) divided by a field element d^2-d-2.

Note that this implies that s is not in the field -- else sqrt(7) would be
in our cubic field too, which is impossible (it would make our cubic field
an extension of Q(sqrt(7)) of degree 3/2).

> [ . . . ] You can rotate this regular heptagon in three-dimensional space with
> three degrees of freedom. No matter how much I rotate it, however, I
> cannot seem to get all the coordinates of all the vertices in the field.
> (Translating the heptagon makes no difference, [ . . . ]
> Even though this is not a rigorous proof that your proof is
> wrong, I am beginning to suspect that there might be a flaw
> in your complicated proof. [ . . . ]

I don't understand why. If you tried hard to find a rotation that works,
and never quite found one, I'd think that this should corroborate
my claim that none exists, not cast doubt on it.

The proof is not all that complicated. Starting from a putative heptagon,
I translate it to put the center at the origin (noting, as you now do too,
that this does not affect the problem's condition); then use a cross product
to find a vector (X,Y,Z) with coordinates in the field whose length is
sqrt(7) times a field element; and finally show that this is impossible
despite the rotational freedom. The last step generalizes the familiar
proof that 7 is not the sum of three rational squares by considering
the possibilities mod 4 and 8. In our settingtThere are more possibilities
than or rational numbers, but it's easy to check by computer. To be sure
it's possible to make coding errors; if you _can_ write 7 = X^2 + Y^2 + Z^2
with X,Y,Z in our cubic field, then my proof is wrong -- and in fact
according to David Ash you can then construct a regular heptagon.
But I expect that the result is correct, because I also found a more
conceptual proof (using a bit more of the structure of p-adic fields)
that works for any unramified extension of Q_2 of odd degree:
<https://people.math.harvard.edu/~elkies/4squares.pdf>
(and checked that the computer proof extends to the first few
cases past d=3, namely 5, 7, and 9).

NDE

[t...@bellefleurbooks.com]

Dear Noam:

As I suspected, I am overlooking something simple. Your proof was that
the heptagon was impossible. I had gotten it backwards and thought
mistakenly that you were proving the heptagon existed. I ran out time
posting the message below, did not have time to read your message
thoroughly or to get back to this website for ten days. I agree. The
heptagon looks impossible. As far as I can tell, your proof is correct.

Sincerely,

Tom

On 2022-04-06 01:54,
competition-corner-p...@googlegroups.com wrote:
> competition-corner-p...@googlegroups.com [1]
> Google Groups [2]
> [2]
>
> Topic digest
> View all topics [1]
>
> * Digest for
> competition-corner-p...@googlegroups.com - 2 updates
> in 1 topic - 1 Update
>
> Digest for competition-corner-p...@googlegroups.com
> - 2 updates in 1 topic [3]
>
> t...@bellefleurbooks.com: Apr 05 08:44PM -0500
>
> Dear Noam:

>
> I had thought that you could show that s is in the field as well as c
> and then make a relatively simple heptagon to demonstrate the
> conjecture. As it turns out, by going down this path, you do not find
> that s is in the field. Rather, as it turns out, you find that
> (7^0.5)s
> is in the field.
>

> As I suggested, you can let k = pi / 14, in which case c = cos 4k and
> s
> = sin 4k. You can make a regular heptagon by letting, for n = 0, 1, 2,
>
> 3, 4, 5, 6, the vertices have the following coordinates:
>
> (x, y, z) = (cos 4kn, sin 4kn, 0).

>
> You can rotate this regular heptagon in three-dimensional space with
> three degrees of freedom. No matter how much I rotate it, however, I
> cannot seem to get all the coordinates of all the vertices in the
> field.

> (Translating the heptagon makes no difference, because if all the
> coordinates of all the vertices of the translated heptagon were in the
>
> field, you could subtract their vector average from them and show that
> a
> regular heptagon with a center at the origin also has coordinates all
> in
> the field.) Even though this is not a rigorous proof that your proof

> is
> wrong, I am beginning to suspect that there might be a flaw in your

> complicated proof. Maybe I am wrong. There might be something simple
> that I am overlooking.
>
> I any case, it turns out that sin k = cos 6k = - cos 8k = 1 - 2 cos^2
> 4k
> = 1 - 2c^2, so sin k is in the field, as I suggested. You can show by
> induction that sin nk is in the field for odd integers n and that cos
> nk
> is in the field for even integers n. Once you know (7^0.5)s is in the
> field, you can show by induction that (7^0.5) sin nk is in the field
> for
> even integers n and that (7^0.5) cos nk is in the field for odd
> integers
> n.
>
> To show that (7^0.5)s is in the field, recall that the sum of the real
>
> components of the seventh roots of unity vanishes, i.e.,
>
> 0 = cos 0 + 2 cos 4k + 2 cos 8k + 2 cos 12k
>
> = 1 + 2c + 2 (2c^2 - 1) + 2 (4c^3 - 3c)
>
> = 8c^3 + 4c^2 - 4c - 1, as David demonstrated.
>
> Therefore, 0 = (8c^3 + 4c^2 - 4c - 1) (- 8c^3 + 4c^2 + 4c - 1)
>
> = 64c^6 - 80c^4 + 24c^2 - 1
>
> = 64 (1 - s^2)^3 - 80 (1 - s^2)^2 + 24 (1 - s^2) - 1
>
> = - 64s^6 + 112s^4 - 56s^2 + 7
>
> = [8s^3 - 4 (7^0.5) s^2 + 7^0.5] [- 8s^3 - 4 (7^0.5) s^2 + 7^0.5].
>
> You can compute - 8s^3 - 4 (7^0.5) s^2 + 7^0.5 numerically, and it is
> not zero, by inspection.
>
> Therefore, 0 = 8s^3 - 4 (7^0.5) s^2 + 7^0.5.
>
> Multiplying both sides by the square root of 7,
>
> 0 = 8 (7^0.5) s^3 - 28s^2 + 7
>
> = 8 (1 - c^2) (7^0.5) s - 28 (1 - c^2) + 7,
>
> so (7^0.5) s is in the field.
>
> Sincerely,
>
> Tom
>
> On 2022-03-28 01:54,
>
> Back to top
>
> You received this digest because you're subscribed to updates for
> this group. You can change your settings on the group membership page
> [4].
> To unsubscribe from this group and stop receiving emails from it send
> an email to
> competition-corner-partici...@googlegroups.com.
>
>
>
>
> Links:
> ------
> [1]
> https://groups.google.com/forum/?utm_source=digest&utm_medium=email#!forum/competition-corner-participant-discussion/topics
> [2]
> https://groups.google.com/forum/?utm_source=digest&utm_medium=email/#!overview
> [3]
> http://groups.google.com/group/competition-corner-participant-discussion/t/b76141b1db6f7165?utm_source=digest&utm_medium=email
> [4]
> https://groups.google.com/forum/?utm_source=digest&utm_medium=email#!forum/competition-corner-participant-discussion/join