Conjeture or theorem?2
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From: Ludovicus on 18 Jul 2010 08:27 Thanks to Tim Little for the answer to my former question on Fibonacci. I found its demonstration in Vorobiev's "Numeros de Fibonacci" Editorial Mir Moscow 1974 Now I have this new question: Is this a Conjecture or a theorem?: "Every odd number can be decomposed in the sum of two numbers such that the sum of their squares is a prime number." Example: 33 = 5 + 29 ; 5^2 + 28^2 = 809 (prime) Ludovicus
From: Tim Little on 18 Jul 2010 09:30 On 2010-07-18, Ludovicus <luiroto(a)yahoo.com> wrote: > Now I have this new question: Is this a Conjecture or a theorem?: > "Every odd number can be decomposed in the sum of two numbers > such that the sum of their squares is a prime number." As far as I know, it is conjecture. It does seem fairly closely related to the Goldbach Conjecture, but I don't immediately see that it implies or is implied by GC. It is known that there are infinitely many primes that are the sum of two squares, so at least it is possible for it to be true. By a terribly inefficient computer search, it does hold at least for odd numbers less than 750 000. - Tim
From: Ludovicus on 18 Jul 2010 18:07 On 18 jul, 09:30, Tim Little <t...(a)little-possums.net> wrote: > On 2010-07-18, Ludovicus <luir...(a)yahoo.com> wrote: > > > Now I have this new question: Is this a Conjecture or a theorem?: > > "Every odd number can be decomposed in the sum of two numbers > > such that the sum of their squares is a prime number." > > As far as I know, it is conjecture. It does seem fairly closely > related to the Goldbach Conjecture, but I don't immediately see that > it implies or is implied by GC. It is known that there are infinitely > many primes that are the sum of two squares, so at least it is > possible for it to be true. > > By a terribly inefficient computer search, it does hold at least for > odd numbers less than 750 000. > > - Tim Yes. Each prime of the form 4n + 1 is the sum of two squares. Then a composite product of primes of that form must also be the sum of two squares. Because (a^2 + b^2)*(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2. Its very probable that the conjecture be true because each odd number 2m + 1, have m decompositions as sum of two distint integers, that is m posibilities that one sum of that squares be a prime. Ludovicus
From: Gerry Myerson on 29 Jul 2010 00:16 In article <14bcfa34-e090-49e2-b7bf-50447bdbe0c9(a)d37g2000yqm.googlegroups.com>, Ludovicus <luiroto(a)yahoo.com> wrote: > Thanks to Tim Little for the answer to my former question on > Fibonacci. > I found its demonstration in Vorobiev's "Numeros de Fibonacci" > Editorial Mir Moscow 1974 > > Now I have this new question: Is this a Conjecture or a theorem?: > "Every odd number can be decomposed in the sum of two numbers > such that the sum of their squares is a prime number." > Example: 33 = 5 + 29 ; 5^2 + 28^2 = 809 (prime) > Ludovicus This is related to http://www.research.att.com/~njas/sequences/A079886 although the particualr question raised here is not raised there. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: hagman on 29 Jul 2010 11:17
On 18 Jul., 14:27, Ludovicus <luir...(a)yahoo.com> wrote: > Thanks to Tim Little for the answer to my former question on > Fibonacci. > I found its demonstration in Vorobiev's "Numeros de Fibonacci" > Editorial Mir Moscow 1974 > > Now I have this new question: Is this a Conjecture or a theorem?: > "Every odd number can be decomposed in the sum of two numbers > such that the sum of their squares is a prime number." > Example: 33 = 5 + 29 ; 5^2 + 28^2 = 809 (prime) > Ludovicus It's at least somewhat likely: Between n and 2 n there are about n/(ln(2n)) primes p = 1 mod 4. Each of these can be uniquely written as p = a^2 + b^2 with a even, b odd. This gives rise to a decomposition k = a + b for some odd number sqrt(n) < k < 2 sqrt(n). Naively, one would expect each k in that range to be hit about 2 sqrt(n)/ln(2n) times |