Proof of LL theorem - Page 5

Numbers · 2006-02-15, 10:04

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.

Kees · 2006-02-15, 10:42

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

Numbers · 2006-02-15, 10:56

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.

alpertron · 2006-02-19, 13:14

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.

Numbers · 2006-02-19, 14:09

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.

mfgoode · 2006-02-19, 15:35

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 · 2006-02-19, 15:41

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

alpertron · 2006-02-19, 16:09

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.

R.D. Silverman · 2006-02-20, 18:03

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.

mfgoode · 2006-02-21, 14:54

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

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2006-02-15, 10:42	#46
Kees Dec 2005 2²×7² Posts	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 * 2pq 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

2006-02-15, 10:56	#47
Numbers Jun 2005 Near Beetlegeuse 2²·97 Posts	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.

2006-02-19, 13:14	#48
alpertron Aug 2002 Buenos Aires, Argentina 2526₈ Posts	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.

2006-02-19, 14:09	#49
Numbers Jun 2005 Near Beetlegeuse 2²·97 Posts	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.