Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2 are the only integer solutions of n! = x^2 + y^2 but is there a proof? R.
RKG wrote:
Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2
are the only integer solutions of
n! = x^2 + y^2
but is there a proof? R.
A number n can be written as a sum of two squares if and only if every prime congruent to 3 mod 4 appears in n's prime factorization with an even exponent. And Erdos proved the following refinement of Bertrand's postulate: the interval from n to 2n always contains both a 1 mod 4 and a 3 mod 4 prime. That'll do it. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 7/3/06, Michael Kleber <michael.kleber@gmail.com> wrote: ...
And Erdos proved the following refinement of Bertrand's postulate: the interval from n to 2n always contains both a 1 mod 4 and a 3 mod 4 prime. That'll do it.
Um ---- n = 0? n = 1? n = 2? By engineer's induction (only slightly modified), this needs to go back to the drawing board! [Neat idea, though.] WFL
--- Fred lunnon <fred.lunnon@gmail.com> wrote:
On 7/3/06, Michael Kleber <michael.kleber@gmail.com> wrote: ...
And Erdos proved the following refinement of Bertrand's postulate: the interval from n to 2n always contains both a 1 mod 4 and a 3 mod 4 prime. That'll do it.
Um ---- n = 0? n = 1? n = 2? By engineer's induction (only slightly modified), this needs to go back to the drawing board! [Neat idea, though.] WFL
"ErdÅs also proved there always exists at least two prime numbers p with n < p < 2n for all n > 6. Moreover, one of them is congruent to 1 modulo 4, and another one is congruent to â1 modulo 4." Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
And Erdos proved the following refinement of Bertrand's postulate: the interval from n to 2n always contains both a 1 mod 4 and a 3 mod 4 prime. That'll do it.
Um ---- n = 0? n = 1? n = 2? By engineer's induction (only slightly modified), this needs to go back to the drawing board!
Yes, sorry, excessively brief: "For all sufficiently large n," and in particular, n>6 will do it. Which explains RKG's handful of solutions perfectly. --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
with an even exponent. And Erdos proved the following refinement of Bertrand's postulate: the interval from n to 2n always contains both a 1 mod 4 and a 3 mod 4 prime.
The Prime Number Theorem implies that given e>0, there exists N such that for all x>N there is a prime between x and (1+e)x. Can this be strengthened to "there is a prime = 1 mod 4 and a prime = 3 mod 4 between x and (1+e)x" ? Gary McGuire
You're asking for a quantitative form of Dirichlet's theorem. There's plenty about it on the web, tho' mostly concerned with consecutive sequences rather than worst-case gaps. WFL On 7/4/06, Gary McGuire <Gary.McGuire@nuim.ie> wrote:
with an even exponent. And Erdos proved the following refinement of Bertrand's postulate: the interval from n to 2n always contains both a 1 mod 4 and a 3 mod 4 prime.
The Prime Number Theorem implies that given e>0, there exists N such that for all x>N there is a prime between x and (1+e)x.
Can this be strengthened to "there is a prime = 1 mod 4 and a prime = 3 mod 4 between x and (1+e)x" ?
Gary McGuire
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No, I don't have a proof, I do have an argument following from a solid conjecture. We know that for n >= 2, there exists a prime p with n < p < 2n. It also appears that for n >= 4, there exists prime p == 3 (mod 4) with n < p < 2n. I cannot prove this myself, but I am sure it is true. This conjecture implies that for n >= 7, there is a prime p == 3 (mod 4) with [n/2] < p <= n. This p occurs with exponent 1 in the prime factorization of n, so that n! cannot be of the form x^2+y^2.
participants (6)
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David Wilson -
Eugene Salamin -
Fred lunnon -
Gary McGuire -
Michael Kleber -
Richard Guy