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 View Poll Results: Regular polygon with ruler & compass, only if n is Power of 2 times 1 or 3 or 5 or 17 0 0% Power of 2 times 1 or prime Fermat Number only 0 0% Power of 2 times 1 or any Fermat Number 1 16.67% Power of 2 times 1 or product of one or more distinct prime Fermat Numbers 3 50.00% Power of 2 times 1 or product of one or more distinct Fermat Numbers 0 0% Power of 2 times 1 or product of one or more distinct or same prime Fermat Numbers 0 0% Power of 2 times 1 or product of one or more distinct or same Fermat Numbers 0 0% Possible for any positive integer value of n 2 33.33% Voters: 6. You may not vote on this poll

 2009-10-26, 08:40 #1 Raman Noodles     "Mr. Tuch" Dec 2007 Chennai, India 100111010012 Posts Regular polygon: ruler & compass possible only if Code: Question: A regular polygon of n sides can be constructed with ruler and compass only if n is ___? 1) Obviously it is possible if n is a power of 2, as you can go on by bisecting angles. 2) And for n = 3, since cos 60° = 0.5, you can easily be able to do so. 3) For n = 5, you have sin 18° = $\sqrt 5 - 1 \over 4$ and so sin 72° = $\sqrt{10+2\sqrt 5} \over 4$. A segment of length $\sqrt 5$ can be constructed by using the hypotenuse of a right angled triangle whose sides are of lengths 1 and 2. 4) Gauss constructed a regular 17-gon by using ruler and compass only. The steps of this construction can be found out at this website. 5) It is obviously possible for power of two times 3 or 5 or 17 because of the fact that you can always bisect these angles, NOT sides of the polygon. Code: For constructing a regular polygon, you are given the center of the polygon and then one of the vertices of the polygon. You must construct all the other sides of the polygon. Obviously if one is constructed, the rest are all trivial! Code: Formally, using ruler and compass means: 1) You can mark a random point on paper 2) You can draw a line or segment or ray between two points 3) You can draw a circle (arc) with center as first point, passing through the second point 4) Mark the intersection of two lines, two circles or a line and circle. 5) Hide any unwanted lines, circles or points to clean up the picture that looks congested. My question: For what values of n will you be able to construct a regular polygon by using ruler and compass only? 1) Is it possible to construct the polygon for prime Fermat numbers such as 257 and 65537? 2) Is it possible to do so for all Fermat numbers such as 4294967297 or 18446744073709551617? If it is possible, then it should be evident for multiples of 641, 274177, 6700417, 67280421310721, etc. 3) Is it possible for product of Fermat numbers, for example, n = 15? 4) Is it possible for powers of Fermat numbers, for example, n = 9? 5) Is it possible for any other values of n, for example n = 7?
 2009-10-26, 09:19 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,433 Posts The answer is easy to find
 2009-10-26, 11:22 #3 Raman Noodles     "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts Let's make the question popular to everyone. If regular polygon with 257 sides and 65537 sides can be constructed, then why not at all for 4294967297 or 18446744073709551617? Why not every Fermat number have a notation like this? $cos \ \frac{\pi}{5} = \frac{\sqrt{10 + 2 \sqrt 5}}{4}$ $cos \ \frac{2\pi}{17} = \frac{-1 + \sqrt {17} + \sqrt{34 - 2 \sqrt {17}} + 2 sqrt{17 + 3 \sqrt {17} - \sqrt{34 - 2 \sqrt {17}} - 2 \sqrt{34 + 2 \sqrt {17}}}}{16}$ It is forming up a sequence, right? It is quite evident that regular 15-gon can be constructed by using equilateral triangle and regular pentagon, by placing their vertices over a circle, and then atleast one vertex of triangle and one vertex of pentagon form the adjacent vertices of the 15-gon. But, is it not at all possible for the powers of Fermat Numbers, such as for n = 9? cos 40° forms a simple root of the cubic equation, with integer coefficients only. 8x3 - 6x + 1 = 0. Last fiddled with by Raman on 2009-10-26 at 11:32
2009-10-26, 11:42   #4
10metreh

Nov 2008

91216 Posts

Quote:
 Originally Posted by Batalov The answer is easy to find

It's possible for all polygons except the diabolical monstragon, which has an indefinite number of sides and tries to eat you before you have finished constructing it.

 2009-10-26, 14:44 #5 wblipp     "William" May 2003 New Haven 44718 Posts Is this a trick question? Classical constructions are done with a compass and straightedge. Substituting a ruler for the straightedge introduces many new possibilities depending on the markings on the ruler. For example, squaring the circle is possible if pi is marked on the ruler.
2009-10-26, 16:31   #6
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

2·5·587 Posts

Quote:
 Originally Posted by wblipp Is this a trick question? Classical constructions are done with a compass and straightedge. Substituting a ruler for the straightedge introduces many new possibilities depending on the markings on the ruler. For example, squaring the circle is possible if pi is marked on the ruler.
I think he meant straightedge. In England a ruler is what we use when we want a straightedge. We very rarely use unmarked rulers.

2009-10-26, 16:33   #7
mdettweiler
A Sunny Moo

Aug 2007
USA (GMT-5)

3×2,083 Posts

Quote:
 Originally Posted by henryzz I think he meant straightedge. In England a ruler is what we use when we want a straightedge. We very rarely use unmarked rulers.
Yes, same here in America. Nowadays it's usually easier to just grab a ruler and ignore the markings when doing a construction.

2009-10-26, 17:05   #8
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

638410 Posts

Quote:
 Originally Posted by Raman Let's make the question popular to everyone. If regular polygon with 257 sides and 65537 sides can be constructed, then why not at all for 4294967297 or 18446744073709551617?
Basically, because the construction relies on the fact that there is a 2^nth root of unity in the field Z/(2^n+1)Z, and this is the case only if n is prime.

I am sure Dr Silverman will jump on me for over-simplifying this, but you can prove that straight-edge and compasses can only construct points whose coordinates satisfy irreducible polynomial equations of order a power of two; so if your point needs to have some coordinate satisfying an irreducible polynomial of order three, you're in trouble.

2009-10-26, 18:34   #9
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by fivemack Basically, because the construction relies on the fact that there is a 2^nth root of unity in the field Z/(2^n+1)Z, and this is the case only if n is prime. I am sure Dr Silverman will jump on me for over-simplifying this, but you can prove that straight-edge and compasses can only construct points whose coordinates satisfy irreducible polynomial equations of order a power of two; so if your point needs to have some coordinate satisfying an irreducible polynomial of order three, you're in trouble.
Sort of. p-1 must have order a power of two because we can only
construct lengths that lie in a simple quadratic extension field of a
length that we can already construct. We can construct any element
of Q. We can only now construct elements lying in a quadratic extension
of an element of Q. Then only in a quadratic extension of that field, etc.
We get a tower of fields, each having index a power of two over ALL of its
sub-fields.

 2009-10-27, 01:32 #10 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9,433 Posts Already last night, browsing away from the (once again, "easy to find") Constructible polygon wiki page, I've reached the page about J.H.Conway (who doubted Hermes' construction) and then, to the delectable 71-degree polynomial for the eponymous constant. Wanted to share its beauty. When I was in high school, I must have read all Gardner's books, translated into Russian and fairly popular in Russia back then, and Conway's name was in each of them, it seems. Last fiddled with by Batalov on 2009-10-27 at 01:38 Reason: ECC for the laned memory
2009-10-27, 09:14   #11
Raman
Noodles

"Mr. Tuch"
Dec 2007
Chennai, India

3×419 Posts

Also I noticed that
$cos \ {2 \pi \over 7}$ is a root of the cubic equation
$8x^3 + 4x^2 - 4x - 1 = 0$

$cos \ {\pi \over 9}$ is a root of the cubic equation
$8x^3 - 6x - 1 = 0$

$cos \ {2 \pi \over 9}$ is a root of the cubic equation
$8x^3 - 6x + 1 = 0$

Further, I know that the roots of the cubic equation $x^3 + qx + r = 0$ are given by
$\sqrt[3]{-{r \over 2}+ \sqrt {{r^2 \over 4}+{q^3 \over 27}}} + \sqrt[3]{-{r \over 2}- \sqrt {{r^2 \over 4}+{q^3 \over 27}}}$,

Here notice that the cube roots of these numbers need not be real at all, they can be complex as well! But always it is true that the complex root will be multiplied by $\omega$ or $\omega^2$ times the real cube root. Where $\omega$, $\omega^2$ are the complex cube roots of unity, being equal to $-{1 \over 2}+{\sqrt 3 \over 2}i$, $-{1 \over 2}-{\sqrt 3 \over 2}i$ respectively.
The roots of the equation $x^2+x+1 = 0$, being derived up right from the equation $x^3 = 1$, or that $x^3 - 1 = 0$, $x^3 - 1 = (x+1)(x^2+x+1)$

and then that any equation of the form $ax^3 + bx^2 + cx + d = 0$ can be converted into the type $y^3 + qy + r = 0$ by substituting y = x + h, and then solving for the value of h, such that the coefficient of the $x^2$ term is 0.

Quote:
 Originally Posted by fivemack Basically, because the construction relies on the fact that there is a $2^{n\ th}$ root of unity in the field $\mathbb{Z}/(2^n+1)\mathbb{Z}$, and this is the case only if n is prime.
For every Fermat Number, there is a polynomial of degree a power of 2, right.
Then, why does it not work out at all for 4294967297?
$\prod_{n=1}^{4294967296} (x - cos \ {n\pi \over 4294967297})$

It is of course, reducible, but the minimal polynomial is of the order $\phi(4294967297)$ = 640 x 6700416 = 4288266240. It is not a power of 2.
But the original polynomial is of the order 4294967296, which means that it has 4294967296 square roots, all of whose surds are of the order being a power of 2 only.
All lengths of this type, can be constructed by using the Pythagoreas Theorem, right? $\sqrt 2$ can be constructed by using the hypotenuse of right angled triangle, containing sides of length 1 and 1. How about the $\sqrt[4] 2$? Hypotenuse of right angled triangle with sides of length 1 and $\sqrt 2$ gives up only $\sqrt 3$!

If any prime Fermat number exists after 65537
"Fermat Prime, maybe the largest" - Brent - why uncertainity? Naturally, it can be only either true or false. If true, then write with confidence, "Fermat Prime, being the largest"!

If any Fermat prime exists after 65537, and it is possible to construct the regular polygon with n sides, where n = this new Fermat Number, then why not is it possible to derive from that for those lower Fermat Numbers?

By the way, is it possible that for any Fermat primes to exist above 65537? It is known till today that the Fermat Numbers $2^{2^n} + 1$, for n = 5 through 32 are composite, as well as for the some other higher values of n. What is the probability? Very low, right? Since all numbers till n = 32 are being tested up so far, if any Fermat prime existed, the majority of the probability should be lying up within this range.

Mainly, I wanted to ask up, what is meant by "Number of Fermat Primes is finite"? This statement looks up right to be ridiculous to me, because finite number of elements to an infinite set to which a bound is not at all known on the upper side, is nonsense. I have read that it is an heuristic argument only, but however, anyway, it is true that, "Number of Fermat Primes is 5" or "There exist no Fermat primes above so and so" are being clearly sensible statements only. But, why is the number of Fermat numbers 'Finite'? Even if there is only one Fermat prime over a very large interval, and then that interval grows up exponentially as n grows larger and larger, this sequence itself will easily count up to infinity. Thus, anyone want to make this statement much clear? "The Number of possible Fermat Primes is only Finite!"?

Last fiddled with by Raman on 2009-10-27 at 09:28

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