

Title: #Primitive pythagorean triples, given hypotenuse Post by Aryabhatta on Feb 4^{th}, 2010, 9:40am Prove/Disprove: Given a positive integer c, the number of primitive pythagorean triples (a,b,c) with c as hypotenuse is either 0 or a power of 2. Note, we only count the triples where a >0, b>0 and c > 0. Primitive pythagorean triples are triples where gcd(a,b) = 1 and a^{2} + b^{2} = c^{2} Example: c = 3 : number of triples = 0. c = 5: Number of triples = 2. (3,4,5) and (4,3,5). 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by SMQ on Feb 5^{th}, 2010, 5:51am Proof by MathWorld (http://mathworld.wolfram.com/PythagoreanTriple.html): see equations (25) and (26). ;D I'm still working on tracing that statement back to elementary principles... SMQ 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by Obob on Feb 5^{th}, 2010, 8:16am I found that page as well, but why it should be true eludes me. 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by rmsgrey on Feb 5^{th}, 2010, 9:08am It's obvious that it has to be even  if a,b,c is a primitive triple, so is b,a,c  it's obvious that no triple can be a,a,c. 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by SMQ on Feb 5^{th}, 2010, 10:00am Alright, so, not so elementary, and still incomplete, but at least based on other "well known" proofs: Consider the factorization of c as 2^{l}p_{1}^{m1}p_{2}^{m2}...p_{r}^{mr}q_{1}^{n1}q_{2}^{n2}...q_{s}^{ns} where each p_{i} is a prime of the form 4x  1 and each q_{i} is a prime of the form 4x + 1 (where x is an integer). We are interested in the number of ways c^{2} = 2^{2l}p_{1}^{2m1}p_{2}^{2m2}...p_{r}^{2mr}q_{1}^{2n1}q_{2}^{2n2}...q_{s}^{2ns} can be represented as the sum of two squares. 1) 2^{2l} cannot be represented by the sum of two squares which are not both divisible by 4. 2^{2l} http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 0 (mod 4). Any square is congruent to 0 or 1 (mod 4), therefore any two squares which sum to 2^{2l} must be congruent to 0 (mod 4) and so divisible by 4. 1a) If l > 0, no primitive Pythagorean triple exists with c as an hypotenuse. By (1), any squares which sum to c^{2} must have 4 as a common factor and therefore are not relatively prime. 2) No odd prime p of the form 4x  1 (where x is an integer) can be represented as the sum of two squares. p http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/equiv.gif 3 (mod 4), but the sum of two squares can only be 0, 1, or 2 (mod 4). 2a) If r > 0, no primitive Pythagorean triple exists with c as an hypotenuse. By (2), any squares which sum to c^{2} must have each p_{i}^{2mi} as a common factor and therefore are not relatively prime. 3) Every prime q of the form 4x + 1 (where x is an integer) can be represented as the sum of two squares. This is Fermat's theorem on sums of two squares (http://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares), first proven by Euler. 3a) No prime q of the form 4x + 1 (where x is an integer) can be represented as the sum two odd squares nor two even squares. This follows trivially from the fact that q is odd. 3b) Every prime q of the form 4x + 1 (where x is an integer) can be represented as the sum of two squares in only one way (up to order and negation). This is known as Thue's Lemma (http://planetmath.org/encyclopedia/ThuesLemma2.html) and an elementary proof of uniqueness can be found at the top of this page (http://planetmath.org/encyclopedia/ProofOfThuesLemma.html). 4) Every power of q can be represented as the sum of two squares. By (3b) there exist some unique positive integers u, v such that u^{2} + v^{2} = q. By the BrahmaguptaFibonacci identity (http://en.wikipedia.org/wiki/BrahmaguptaFibonacci_identity) (hereafter BF identity), (u^{2}  v^{2})^{2} + (2uv)^{2} = q^{2}. Proof for arbitrary powers is by induction in the BF identity. 4a) Every representation of a power of q as a sum of two squares is generated by iteratively applying the BF identity. I do not yet have a proof of this step. 4b) Every power of q can be represented as the sum of two coprime squares in exactly one way (up to order and negation). I believe this can be proven by induction using (4a) and divisibility by lower powers (showing that one of the two solutions generated by iteratively applying the BF identity is always divisible by q), but the details are morethanalittle messy. 5) For q_{i}^{2ni} and q_{j}^{2nj}, the solutions generated by applying the BF identity to u_{ni}^{2} + v_{ni}^{2} and u_{nj}^{2} + v_{nj}^{2} are distinct. The BF identity yeilds (u_{ni}u_{nj}  v_{ni}v_{nj})^{2} + (u_{ni}v_{nj} + v_{ni}u_{nj})^{2} and (u_{ni}u_{nj} + v_{ni}v_{nj})^{2} + (u_{ni}v_{nj}  v_{ni}u_{nj})^{2}. Of these, u_{ni}u_{nj}  v_{ni}v_{nj} and u_{ni}u_{nj} + v_{ni}v_{nj} clearly have the same parity but are distinct, while by (3a) the other terms have the opposite parity. By symmetry all four terms must be distinct. 5a) Both solutions found in (5) are primitive. I have no proof of this step: the result of (4b) does not suffice. 5b) Continuing in the same manner, the product of q_{i}^{2ni}, q_{j}^{2nj}, and q_{k}^{2nk} has four distinct primitive solutions, etc. I have no proof for this step, neither for distinctness nor for primitiveness. 5c) All integral solutions to the original equation a^{2} + b^{2} = c^{2} can be enumerated in the manner of (5b). And, again no proof. 6) Therefore, either a^{2} + b^{2} = c^{2} = 2^{2l}p_{1}^{2m1}p_{2}^{2m2}...p_{r}^{2mr}q_{1}^{2n1}q_{2}^{2n2}...q_{s}^{2ns} has no primitive solutions if either l > 0 or r > 0, by (1a) and (2a), or there are 2^{s  1} distinct (up to order and negation) primitive solutions by (4b) and (5b). SMQ Edit: found a few more holes... Edit 2: changes (3b) to reference Thue's Lemma Edit 3: found one more hole, added (5c)... 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by Aryabhatta on Feb 6^{th}, 2010, 11:49am 4a and 4b are true. 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by SMQ on Feb 8^{th}, 2010, 5:48am on 02/06/10 at 11:49:48, Aryabhatta wrote:
And given those, if (5c) is true (which seems very likely), (5a) and (5b) must be true in order to obtain the "correct" answer. I'm confident in the reasoning, I just haven't yet found or devised proofs for the later steps... SMQ 

Title: Re: #Primitive pythagorean triples, given hypotenu Post by Aryabhatta on Feb 8^{th}, 2010, 7:40am on 02/08/10 at 05:48:50, SMQ wrote:
I think 5a,b,c are all true too. 

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