Re: twin prime conjecture

Hi Gerry,

I must admit that I am not certain that I have a proof so it may not be
worth your time to review it. If on the other hand, you would like to
help me find the flaw or help review it per chance that it is has any
validity, I will greatly appreciate it.

In the spirit of doing everything I can to encourage a number theorist
to help me review my proof, I will provide the answers to most of your
questions. I will try to give you the answers to the last questions
tomorrow or later this week. :-)

1. Show that if integers x and y can be expressed as a sum of two
squares of integers, then so can xy.

How's this:

(a^2 + b^2)(c^2 + d^2) =

= (a^2)(c^2) + (a^2)(d^2) + (b^2)(c^2) + (b^2)(d^2) =

= (a^2)(c^2) + 2abcd + (b^2)(d^2) + (a^2)(d^2) -2abcd +(b^2)(c^2) =

= (ac + bd)^2 + (ad - bc)^2

2. Prove that the cube root of 36 is irrational.

(a) Assume that that there exists x,y such that the cube root of 36 is
rational so that (x/y)^3 = 36

(b) We can assume that gcd(x,y)=1 since if they didn't we could divide
the common factor from (x/y). [All fractions can be reduced to a form
of gcd(x,y)=1 5/10 becomes 1/2 etc.]

(c) This gives us x^3 = 36(y^3)

(d) x is clearly even since 36 divides x^3.

(e) There exists x' such that x=2x' and we have (2x')^3 = 8x^3 =
36(y^3)

(f) We divide 4 from both sides to get: 2x^3 = 9(y^3) and we have a
contradiction since 2 must divide y which is impossible since
gcd(x,y)=1.

3. Prove that the base-10 logarithm of 2 is irrational.

Assume that log10(2) is rational for some positive integers m,n so that
log10(2)=m/n

It follows that 10^(m/n)=2

Raise each side ot the n power so that 2^n = 10^m

We now have 2^n = 2^m*5^m = 2^(n-m) = 5^m which is impossible.

4. Find the smallest positive integer n such that 2 n is the
square of an integer, 3 n is the 3rd power of an integer, and 5 n is
the 5th power of an integer.

This one was fun! Thanks very much since I had not seen it before. I
will skip some steps because the solution is more fun than explaining
how I came up with it.

The important insight is that there exists x,y,z such that:

(1/2)x^2 = (1/3)y^3 = (1/5)z^5 so that we have:

15x^2 = 10y^3 = 6z^5

The solution comes from realizing that we need to organize 2,5,3 so
that they meet the constraints. Lots of fun.

The solution that I came up with is:

z = 5^5*3^4*2^3
y = 5^8*3^7*2^5
x = 5^12*3^10*2^8

n = (1/2)x^2 = 5^24*3^20^2^15

The important point is that the number of powers of 5,4,2 must meet the
following constraints where (x',y',z' are the variables for x,y, and z)

Number of 5's = 2x'+1 = 3y'+1 = 5z'
Number of 3's = 2x'+1 = 3y' = 5z'+1
Number of 2's = 2x' = 3y'+1 = 5z'+1

5. The Fibonacci numbers are given by F_0 = 0, F_1 = 1, and
F_n = F_(n - 1) + F_(n - 2) for n = 2, 3, ... (so they go
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...). Let phi be the
number (1 + sqrt 5) / 2. Prove that
phi^(n - 1) < F_(n + 1) < phi^n if n > 1.

In other words, prove Binet's Formula. If you feel I am cheating here,
I am glad to rework it (but only if you really will review my proof and
send detailed feedback).

Here's a summary, I took from Dr. Math (I am always glad to promote
this wonderful web site!):

http://mathforum.org/library/drmath/view/52717.html

The Principle of Mathematical Induction states that if a certain
statement that depends on n is true for n = 0, and if its truth for
n = k implies its truth for n = k+1, then the statement is true for
all integers n >= 0.

There is an equivalent form, which appears superficially to be
different. It states that if a certain statement that depends on n is
true for n = 0, and if its truth for all n <= k implies its truth for
n = k+1, then the statement is true for all integers n >= 0.

To apply this Principle in either form is to prove the statement "by
induction."

Binet's formula is

F(n) = (a^n-b^n)/(a-b).

Here F(n) is the nth Fibonacci number, defined by

F(0) = 0,
F(1) = 1,
F(n) = F(n-1) + F(n-2), for all n >= 2.

The quantities a and b are defined by the formulas

a = (1+sqrt[5])/2,
b = (1-sqrt[5])/2 = -1/a.

They are the two roots of the quadratic equation x^2 - x - 1 = 0.
(Check this!)

To prove Binet's Formula, first show that it is true for n = 0 and
n = 1. Then for any k >= 1, assume it is true for all n <= k (in
particular for n = k and n = k-1), so that

F(k) = (a^k-b^k)/(a-b),
F(k-1) = (a^[k-1]-b^[k-1])/(a-b).

Now add these two equations together. The left-hand side becomes
F(k+1), according to the recursion defining the Fibonacci numbers.
Rearrange the right-hand side into the form

F(k+1) = (a^k+a^[k-1]-b^k-b^[k-1])/(a-b),
= (a^[k-1]*[a+1]-b^[k-1]*[b+1])/(a-b).

Now use the facts that a + 1 = a^2 and b + 1 = b^2, because a and b
are the roots of x^2 - x - 1 = 0. Then the above expression will
simplify into the form of Binet's Formula for n = k+1.

That establishes the hypotheses of the second form of the Principle of
Mathematical Induction. The conclusion of the Principle must therefore
hold, and Binet's Formula is true for all integers n >= 0.

-Larry


Gerry Myerson wrote:
In article <1160515285.625915.191680@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"larry.freeman@xxxxxxxxx" <larry.freeman@xxxxxxxxx> wrote:

Hi Everyone,

To make it easier for people to follow my argument, I have put together
an informal version of the proof that reduces the main argument to 1
page:

http://proposedproofs.blogspot.com/2006/10/twin-prime-conjecture-informal-proo
f.html

You can still find the complete proof here:

http://proposedproofs.blogspot.com/2006/10/proposed-proof-twin-prime-conjectur
e.html

-Larry


larry.freeman@xxxxxxxxx wrote:
Hi Math Experts,

My name is Larry Freeman and I am a math amateur. I have a draft of a
proposed proof of the twin primes conjecture which can be found here:
http://proposedproofs.blogspot.com/2006/10/proposed-proof-twin-prime-conject
ure.html

I would greatly appreciate it if someone who is an expert in number
theory could take a look at it and explain to me where I went wrong.
This solution came too easily to me for it to be correct.

I've spent close to 2 weeks time on the proof and the argument seems
very solid to me. It is based on very elementary number theory. The
proof is only 9 pages.

Before I spent a second looking at your write-up, you have to
convince me that you can actually do some number theory. Here
are 5 problems for you, problems that would not be out of place
in an intro number theory course. If you can't do them, there
isn't the slightest chance that you can do something as hard
as the twin prime conjecture.

Good luck. And please don't top-post.

1. Show that if integers x and y can be expressed as a sum of
two squares of integers, then so can xy.

2. Prove that the cube root of 36 is irrational. Prove that
the base-10 logarithm of 2 is irrational.

3. Find the smallest positive integer n such that 2 n is the
square of an integer, 3 n is the 3rd power of an integer, and
5 n is the 5th power of an integer.

4. The Fibonacci numbers are given by F_0 = 0, F_1 = 1, and
F_n = F_(n - 1) + F_(n - 2) for n = 2, 3, ... (so they go
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...). Let phi be the
number (1 + sqrt 5) / 2. Prove that
phi^(n - 1) < F_(n + 1) < phi^n if n > 1.
Prove that F_1 + F_2 + F_3 + ... + F_n = F_(n + 2) - 1.

5. Find, with proof, an irreducible polynomial that has
sqrt 2 + sqrt 3 as a root.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)

.

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