20060215, 10:04  #45  
Jun 2005
Near Beetlegeuse
388_{10} Posts 
I am obviously completely thick, because I still don't get it.
Quote:
Let , , form the Pythagorean triple . You say that it follows that from which we get that which factors as either or . So, . You say . or gives us , or . or gives us , or . I don't see a 9 in there, do you? Then you say, . or gives us , or . 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 then , 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 and and construct a Pythagorean triple in which it is true. But it does not "follow" from three integers being a triple. Quote:
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. 

20060215, 10:42  #46 
Dec 2005
2^{2}×7^{2} 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^2q^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^2q^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 
20060215, 10:56  #47 
Jun 2005
Near Beetlegeuse
184_{16} Posts 
Aha!.
so what your saying is that it isn't It is in fact . Thank you kees for that clarification. No doubt someone will now accuse me of being pedantic. 
20060219, 13:14  #48 
Aug 2002
Buenos Aires, Argentina
2510_{8} 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. 
20060219, 14:09  #49 
Jun 2005
Near Beetlegeuse
2^{2}×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: ..................(A) .......(B) When we see them together it is immediately obvious that the in (A) is equivalent to the in (B). I, however, failed to make this connection. So, when you said that 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. 
20060219, 15:35  #50  
Bronze Medalist
Jan 2004
Mumbai,India
804_{16} Posts 
Proof of LL theorem
Quote:
Mally 

20060219, 15:41  #51  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Quote:
Its not good to give in to Self Pity. 'Beware the ides of March' Mally 

20060219, 16:09  #52  
Aug 2002
Buenos Aires, Argentina
2^{3}×13^{2} Posts 
Quote:


20060220, 18:03  #53  
Nov 2003
2^{2}×5×373 Posts 
Quote:
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. 

20060221, 14:54  #54  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Proof of LL theorem
Quote:
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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