Re: Digest for competition-corner-participant-discussion@googlegroups.com - 2 updates in 1 topic

[t...@bellefleurbooks.com]

Dear David and Istvan:

I had to go back to work. I don't have so much time anymore.

As you note, David, 8c^3 + 4c^2 - 4c - 1 = 0. You can let k = pi/14, so
c = cos 4k, and s = sin 4k. You can show that sin k is in the field and
express it as a cubic polynomial of c. Thus, s is in the field too.

Is it not enough, then, to let x = c, y = s and z = 0?

--Tom

On 2022-03-26 01:54,
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> * Problem suggested by David McCooey - 2 Updates
>
> Problem suggested by David McCooey [3]
>
> Istvan Lauko <igl...@gmail.com>: Mar 25 10:29AM -0700
>
> I am relaying a problem from David
>
> here is what he wrote:
>
> About me: In 1980, I went on a mathematics competition trip organized
> by George Berzsenyi and the Texas ARML. It was a very rewarding
> experience, but I entered university in the fall and did not keep in
> touch with George. A couple weeks ago, I contacted George to see
> if he'd be interested in a math problem I had been working on.
> I don't know whether this problem has been solved already.
>
> Here's the problem:
> Find coordinates for a regular heptagon in 3D Euclidean space where
> all 3 components (x,y,z) of all 7 coordinates are elements of the same
> cubic field, or prove that it can't be done.
>
> The problem can be stated equivalently, with more algebra, as follows:
> Find a unit vector (x,y,z) that allows all 9 entries of the following
> rotation
> matrix to be elements of the same cubic field:
>
> (1-c)*x^2 + c (1-c)*x*y - z*s (1-c)*x*z + y*s
> (1-c)*x*y + z*s (1-c)*y^2 + c (1-c)*y*z - x*s
> (1-c)*x*z - y*s (1-c)*y*z + x*s (1-c)*z^2 + c
>
> where:
>
> c = cos(2*pi/7)
> s = sin(2*pi/7)
>
> or prove that no such vector exists.
>
> When I contacted George, it had occurred to me that this problem was
> similar to competition math problems, but the notion of a field is
> usually
> beyond high school mathematics, at least in 1980 USA. However, there
> are always students with exceptional talent and knowledge, and one of
> them may be able to solve this problem.
>
> As I told George, I have not fully explored the number-theoretic
> aspects
> of this problem, so a simple solution or disproof may yet exist.
> I have been pursuing it in my off time, mostly using computer
> searches.
>
> Best regards,
> David McCooey
>
> David Ash <david_...@yahoo.com>: Mar 25 05:49PM -0700
>
> I don't have a complete solution, but some thoughts:
>
> By taking differences we determine that x*s and y*s must be in the
> field,
> and we similarly determine by taking sums that (1-c)*x*y must be in
> the
> field. Taking ratios, (1-c)/(s^2)= 1/(1+c) must be in the field, so c
> must
> be in the field. (If either x=0 or y=0, this would involve a division
> by
> zero, but in that case we can deduce more trivially that c must be in
> the
> field).
>
> As c is a root of the cubic 8x^3+4x^2-4x-1=0, the field must be
> precisely
> Q[c] if a solution exists. As x*s, y*s, and z*s are all in the field,
> let
> x=X/s, y=Y/s, and z=Z/s where X,Y, and Z are all in the field. In this
> case
> it can be shown that all nine elements must be in the field. So a
> necessary
> and sufficient condition for (x,y,z) to be a solution is for x*s, y*s,
> and
> z*s to all be in the field.
>
> In this case x^2+y^2+z^2=1 so X^2+Y^2+Z^2=s^2=1-c^2. So a necessary
> and
> sufficient condition for a solution to exist is to find X,Y,Z in Q[c]
> with
> X^2+Y^2+Z^2=1-c^2.
>
> That's as far as I've gotten.
>
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[David Ash]

Thanks!

There's one step in your reasoning that I'm missing. You say that you can show that s = sin 4k is in the cubic field. Now s = 4 sin k cos k cos 2k.

I'm convinced that sin k is in the field. sin k = sin (pi/14) = - cos (4*pi/7) = 1-2*c^2

I'm also convinced that cos 2k is in the field. cos 2k = cos (pi/7) = - cos (pi-(pi/7)) = - cos (6*pi/7) = 3c-4c^3

To show that s is in the field, it remains to be shown that cos k is in the field. However, I'm not quite so convinced that cos k is in the field. Can you help me out with how you're deriving that?

[Elkies, Noam]

<competition-corner-p...@googlegroups.com> writes:

> [ . . . ] Problem suggested by David McCooey

> [ . . . ]

> Find coordinates for a regular heptagon in 3D Euclidean space where
> all 3 components (x,y,z) of all 7 coordinates are elements of the same
> cubic field, or prove that it can't be done.

That's a nice problem, which I haven't seen before.
It looks like such a heptagon is impossible. My proof follows.

It is more convenient to work with the algebraic integer

d = 2 c = 2 cos(2 Pi/7) = z + 1/z

where z is a 7th root of unity; the minimal equation of d is
d^3 + d^2 - 2*d - 1 = 0. David Ash already showed that
if such a heptagon exists then the field F must be QQ(c),
which is also QQ(d), and that 1-c^2 (equivalently,
4(1-c^2) = 4-d^2) must be a sum of three squares in F.
This turns out to be impossible, for much the same reason that
7 is not a sum of three rational squares: Check that if
X,Y,Z are algebraic integers in F then X^2 + Y^2 + Z^2
(i) cannot be -w^2 mod 8 for any algebraic integer w
that's not a multiple of 2, and (ii) can only be a multiple of 4 if
each of X,Y,Z is a multiple of 2. Then if X^2 + Y^2 + Z^2 = 7 T^2
for some algebraic integers X,Y,Z,T in F with T nonzero
then we get a contradiction by repeated applying (ii) until
finding an equivalent solution with T not divisible by 2,
at which point (i) shows that this is impossible because
4-d^2 is congruent to -(2d^2-d+2)^2 mod 8. [We say that
two algebraic numbers are "congruent mod m" iff they differ by
c*m for some _algebraic_ integer c.]

Some further remarks:

@ David Ash outlined a proof of the equivalence
"heptagon exists <==> 1-c^2 = X^2 + Y^2 + Z^2 in F",
but since we need only the "==>" direction
a simpler proof suffices. Call the cubic field K.
The center of the heptagon has coordinates in K
(it's the average of the coordinates of the vertices),
so we might as well translate to put the center at the origin.
Then all the vertices have the same length, say sqrt(N)
for some nonzero N in K. If v and v' are adjacent vertices
then the dot product v.v' is c N; since v.v' and N are both in K,
so is c, whence K = QQ(c) = F. The cross product of v and v'
is a vector of length s / N; thus s^2 / N^2 = (4-d^2) / (2N)^2
is a sum of three squares in F, whence so is 4-d^2.

@ It so happens that (4-d^2) (d^2-d-2)^2 = 7,
so that (i) above is even more clearly a generalization of
the familiar fact that 7 is not the sum of three rational squares.

@ I checked (i) and (ii) by having the computer run through
all possible of residue classes in ZZ[c] mod 4 or 8.
(First get a list of all possible squares mod 8;
then sort, remove duplicates, and loop over triples in
the resulting duplicate-free list.) I think that I can
give a proof that generalizes this to all fields QQ(x)
generated by a root of some odd-degree polynomial in ZZ[x]
that remains irreducible mod 2 -- but this may be beyond
the scope of the competition-corner-participant list.

@ A regular 9-gon *can* be placed in Euclidean space
with all coordinates in the same cubic field, necessarily
the field generated by 2*cos(2*Pi/9), which is a root of
t^3 - 3*t + 1 = 0. Indeed we may put three vertices at
an equilateral triangle such as (0,1,-1), (-1,0,1), (1,-1,0),
then fill in the remaining 6 vertices as QQ[t]-linear combinations.
Here if we imitate our analysis for the heptagon we end up with
4-t^2 being a sum of three squares, which it is because
4-t^2 = (2t^2+t-4) / 3.

[David Ash]

Noam,

Thanks for this proof! I'm not surprised to see that some kind of sum of squares argument was needed to nail it, given that the problem seemed to reduce to showing that a specific element of a specific field is not the sum of three squares in that field, and definitely appreciate your work on getting to that point!

I'm kind of curious, though, about working out the details in a couple of steps that you've sketched. For example, you state "if X,Y,Z are algebraic integers in F then X^2 + Y^2 + Z^2...can only be a multiple of 4 if each of X,Y,Z is a multiple of 2." I'm wondering how to show this at the next level of detail. We can certainly write X=2X_1+X_2 where X_1 and X_2 are algebraic integers and X_2 is in the set S={0,1,d,d^2,1+d,1+d^2,d+d^2,1+d+d^2}. Then X^2=4(X_1^2+X_1*X_2)+X_2^2. So if X^2+Y^2+Z^2 is a multiple of 4, so also is X_2^2+Y_2^2+Z_2^2. We can reduce the square of each element in S to a quadratic in d and then, by brute force, check out all possible sums of three of them, showing that the sum can only be a multiple of 4 if X_2=Y_2=Z_2=0. This seems a bit tedious to do, though. Do you (or anyone) know of a more elegant approach?

[George Berzsenyi]

Thanks to Michael for the proposal of this problem, to István for uploading it to this website, and to David and Noam for solving it.  It was indeed an interesting problem, deserving a nice solution.

George 

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• Problem suggested by David McCooey - 2 Updates

Problem suggested by David McCooey

Istvan Lauko <igl...@gmail.com>: Mar 25 10:29AM -0700 I am relaying a problem from David   here is what he wrote:       About me: In 1980, I went on a mathematics competition trip organized by George Berzsenyi and the Texas ARML. It was a very rewarding experience, but I entered university in the fall and did not keep in touch with George. A couple weeks ago, I contacted George to see if he'd be interested in a math problem I had been working on. I don't know whether this problem has been solved already.   Here's the problem:

Find coordinates for a regular heptagon in 3D Euclidean space where all 3 components (x,y,z) of all 7 coordinates are elements of the same cubic field, or prove that it can't be done.

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--

Dr. George Berzsenyi  

1818 East Shorewood Blvd., #310 

Shorewood, WI 53211-2539, U.S.A.

Mobile: 414-418-0456 (George / Gyuri)

Website: www.berzsenyifamily.net

Skype name: george.berzsenyi

[t...@bellefleurbooks.com]

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,

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> competition-corner-p...@googlegroups.com - 2 updates
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>
> "Elkies, Noam" <elk...@math.harvard.edu>: Mar 27 08:50AM -0400

>
>> Find coordinates for a regular heptagon in 3D Euclidean space where
>> all 3 components (x,y,z) of all 7 coordinates are elements of the
> same
>> cubic field, or prove that it can't be done.
>

> David Ash <david_...@yahoo.com>: Mar 27 08:58AM -0700

>
> Noam,
>
> Thanks for this proof! I'm not surprised to see that some kind of sum
> of
> squares argument was needed to nail it, given that the problem seemed
> to
> reduce to showing that a specific element of a specific field is not
> the
> sum of three squares in that field, and definitely appreciate your
> work on
> getting to that point!
>
> I'm kind of curious, though, about working out the details in a couple
> of

> steps that you've sketched. For example, you state "*if X,Y,Z are

> algebraic
> integers in F then X^2 + Y^2 + Z^2...can only be a multiple of 4 if
> each of

> X,Y,Z is a multiple of 2*." I'm wondering how to show this at the next

>
> level of detail. We can certainly write X=2X_1+X_2 where X_1 and X_2
> are
> algebraic integers and X_2 is in the set
> S={0,1,d,d^2,1+d,1+d^2,d+d^2,1+d+d^2}. Then
> X^2=4(X_1^2+X_1*X_2)+X_2^2. So
> if X^2+Y^2+Z^2 is a multiple of 4, so also is X_2^2+Y_2^2+Z_2^2. We
> can
> reduce the square of each element in S to a quadratic in d and then,
> by
> brute force, check out all possible sums of three of them, showing
> that the
> sum can only be a multiple of 4 if X_2=Y_2=Z_2=0. This seems a bit
> tedious
> to do, though. Do you (or anyone) know of a more elegant approach?
>
> On Sunday, March 27, 2022 at 5:51:19 AM UTC-7 elkies wrote:
>

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