mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2006-02-15, 10:04   #45
Numbers
 
Numbers's Avatar
 
Jun 2005
Near Beetlegeuse

38810 Posts
Default

I am obviously completely thick, because I still don't get it.

Quote:
Originally Posted by Alpertron
You set arbitrarily the value y=3, that does not give integer values for p and q.
Well actually no I didn't arbitrarily set y=3. I just used the first Pythagorean triple I thought of; y = 3, x = 4, z = 5 and showed that if (x, y, z) are a Pythagorean triple then y^{2} does not have to be of the form 2pq. Since you seem to be under the impression that I chose this deliberately, I will now choose y = 4 (the other possible definition of y from this same triple) and show exactly the same thing.

Let \large x = 3, \large y = 4, \large z = 5 form the Pythagorean triple \large 9\,+\,16\,=\,25.

You say that it follows that \large y^{2} = 2pq from which

we get that \large pq = 8 which factors as

either \large 1 \times 8 or \large 2 \times 4.

So, \large (p,\,q) = (1,\,2,\,4,\,8).


You say \large x^{2} = p^{2} - q^{2}.

\large p\,=\,1,\;q\,=\,8 or \large p\,=\,8,\;q\,=\,1 gives us \large 1 - 64 = -63, or \large 64 - 1 = 63.

\large p\,=\,2,\;q\,=\,4 or \large p\,=\,4,\;q\,=\,2 gives us \large 4 - 16\,=\,-12, or \large 16 - 4\,=\,12.

I don't see a 9 in there, do you?

Then you say, \large z^{2}\,=\,p^{2} + q^{2}.

\large p\,=\,1,\;q\,=\,8 or \large p\,=\,2,\;q\,=\,4 gives us \large 1 + 64\,=\,65, or \large 16 + 4\,=\,20.

I don't see a 25 in there. Do you?

So, using the very first Pythagorean triple I could think of I have shown that it definitely does not follow that if \large x^{2} + y^{2} = z^{2} then y^{2} = 2pq, completely irrespective of which term you call y.

So there is no logical sense in which you can claim that it follows. Because, it does not.

You might be able to start with p and q and construct a Pythagorean triple in which it is true. But it does not "follow" from three integers being a triple.

Quote:
Originally Posted by Alpertron
Have you ever seen a proof by contradiction?
Formally, no. But I have seen things like the proof of the infinity of the primes. A proof by contradiction starts by assuming a statement P to be true. Then it shows that this leads to a contradiction and the assumption is that therefore P is not true.

My understanding is, however, that the route to the contradiction has to consist of one or more logically sound steps. I am not yet in a position to agree that your first step is logically sound.
Numbers is offline   Reply With Quote
Old 2006-02-15, 10:42   #46
Kees
 
Kees's Avatar
 
Dec 2005

22×72 Posts
Default

Before we rewrite history by saying that pythagorean triplets do not work...

a^2+b^2=c^2, gcd(a,b,c)=k

then

a = k * (p^2-q^2)
b = k * 2*p*q
c = k * (p^2+q^2)

in the simple case

3^2+4^2=5^2

we get

3 = p^2-q^2
4 = 2pq

so p=2 and q=1
which leaves

p^2+q^2 = 2^2+1^2 = 5 = c

This is not a demonstration, but just shows that the pythagorean triplets are not completely off
Kees is offline   Reply With Quote
Old 2006-02-15, 10:56   #47
Numbers
 
Numbers's Avatar
 
Jun 2005
Near Beetlegeuse

18416 Posts
Default

Aha!.

so what your saying is that it isn't \large y^{2}\,=\,2pq

It is in fact \large y\,=\,2pq.

Thank you kees for that clarification.

No doubt someone will now accuse me of being pedantic.
Numbers is offline   Reply With Quote
Old 2006-02-19, 13:14   #48
alpertron
 
alpertron's Avatar
 
Aug 2002
Buenos Aires, Argentina

25108 Posts
Default

But 3^2 and 4^2 are not fourth powers!!!

If x^4+y^4 = z^2 it should be clear that x^2 = a, y^2 = b, so y^2 = 2pq.
alpertron is offline   Reply With Quote
Old 2006-02-19, 14:09   #49
Numbers
 
Numbers's Avatar
 
Jun 2005
Near Beetlegeuse

22×97 Posts
Default

Alpertron,

Everything you have said is absolutely correct.
All that happened was that I failed to make a simple connection between two statements you made in post #22.

Lets look at two simple identities:

\large x^{2}\,+\,y^{2}\,=\,z^{2}..................(A)
\large (x^{2})^{2}\,+\,(y^{2})^{2}\,=\,(z^{2})^{2}.......(B)

When we see them together it is immediately obvious that the \large y in (A) is equivalent to the \large y^{2} in (B).

I, however, failed to make this connection. So, when you said that \large y^{2}=2pq I was trying to derive this from (A), but should have been looking at (B).

In other words, I am a completely stupid idiot and you are very patient gentleman.

Thank you.
Numbers is offline   Reply With Quote
Old 2006-02-19, 15:35   #50
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

80416 Posts
Thumbs down Proof of LL theorem

Quote:
Originally Posted by mfgoode
If Numbers is not content then try this URL.
http://homepages.cwi.nl/~dik/mathematics/jsh2.html
Mally
My entire post on PYthagorean Triplets has been erased but I have it saved and wonder if I should repost it.
Mally
mfgoode is offline   Reply With Quote
Old 2006-02-19, 15:41   #51
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

22·33·19 Posts
Cool

Quote:
Originally Posted by Numbers
Alpertron,

In other words, I am a completely stupid idiot and you are very patient gentleman.

Thank you.
I doubt you are as you are a skilled sophist.
Its not good to give in to Self Pity. 'Beware the ides of March'
Mally
mfgoode is offline   Reply With Quote
Old 2006-02-19, 16:09   #52
alpertron
 
alpertron's Avatar
 
Aug 2002
Buenos Aires, Argentina

23×132 Posts
Default

Quote:
Originally Posted by mfgoode
My entire post on PYthagorean Triplets has been erased but I have it saved and wonder if I should repost it.
Mally
Please read this thread about posts being lost. I would recommend you to post again.
alpertron is offline   Reply With Quote
Old 2006-02-20, 18:03   #53
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

22×5×373 Posts
Default

Quote:
Originally Posted by TravisT
I'm halfway through my abstract algebra series (Up through chapter 6 in Artin's Algebra, if you're familiar with it) I know Lagrange's theorem, however, I don't think we've gone over multiplicative sub-groups of a finite field just yet. Would you mind going over the proof with some pointers of where I should do my homework to understand the bits I don't know?

Sure. All you need to know is that for a prime p, GF(p^2) has
a multiplicative subgroup of order (p+1). Look at the elements of
GF(p^2) whose inverse equals their conjugate. Now just find an
element of full order in this group, and show that x_{n+1} = x_{n}^2 - 2
is just doing exponentiation in this group in disguise.
R.D. Silverman is offline   Reply With Quote
Old 2006-02-21, 14:54   #54
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

22×33×19 Posts
Thumbs up Proof of LL theorem

Quote:
Originally Posted by mfgoode
My entire post on PYthagorean Triplets has been erased but I have it saved and wonder if I should repost it.
Mally
Resending my post.
As a footnote to this debate I append the following facts on Pythagorean Triplets (Py. Trips.)
1) Pythagorean Method: For all odd numbers (no.s)
If m is an odd natural no. then
[(m^2 + 1)/2]^2 = [(m^2 – 1)/2] ^2 + m^2
Eg: If m = 17, we get 145^2 =144^2 + 17^2

2)Plato’s method: For any natural no.
(m^2 + 1)^2 = (m – 1) ^2 + (2m)^2 where m is a natural no.
Note: Since (m^2 + 1) and (m^2 – 1) differ by 2 in this formula those no.s differing by 1 cannot be given eg: 7, 24 , and 25 as 24 and 25 differ by 1

3)Euclid’s Method : If x and y are integers and if
a = x^2 – y^2; b = 2xy ; c =x^2 + y^2 t hen a, b, c, are integers such that
a^2 + b^2 = c^2.
These formulae should generate all possible Py. Trips.
I hope this clears the doubt and debate.
Mally
mfgoode is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Proof of Fermat's Last Theorem McPogor Miscellaneous Math 18 2007-10-19 11:40
help with a proof vtai Math 12 2007-06-28 15:34
Proof (?!) that RH is false? bdodson Lounge 6 2007-03-19 17:19
A proof with a hole in it? mfgoode Puzzles 9 2006-09-27 16:37
A Second Proof of FLT? jinydu Math 5 2005-05-21 16:52

All times are UTC. The time now is 08:39.

Sat May 8 08:39:26 UTC 2021 up 30 days, 3:20, 0 users, load averages: 3.32, 3.33, 3.52

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.