Re: Probable Prime Number
- From: mjtg@xxxxxxxxxxxxx (M.J.T. Guy)
- Date: 28 Sep 2006 12:58:52 GMT
In article <17937020.1159429765953.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
bassam king karzeddin <bassam@xxxxxxxxxx> wrote:
Dear All
I would like to introduce the following conjecture about prime
factorization without having a proof or a counter example or a reference.
The Conjecture:
"If, n = (p^2 + q^2) / 2, where, (p, q) are odd prime numbers,
then, (n) has at most three distinct odd prime numbers factors"
Obviously false. Instead we have:
Given any set of primes p_1, ... p_k each of the form 4a+1, we can find
infinitely many n = (p^2 + q^2)/2 with p, q prime and n divisible by all the
p_i.
Proof: First we find a solution without the condition that p, q be prime.
We can express 2*p_i as a sum of squares, so we can find a suitable n
for each primne separately. If (p, q) gives a solution, so does
(p + a*p_i, q + b*p_i) for all (a, b). So we can combine the
individual solutions using the Chinese Remainder Theorem to find an
n divisible by S = p1*p2*...*p_k.
If (p, q) is a solution, so is (p + a*S, q + b*S), so we can find infinitely
many solutions with (p, q) prime, since there are infinitely many
primes in each of the arithmetic series p + a*S and q + b*S.
Mike Guy
.
- References:
- Probable Prime Number
- From: bassam king karzeddin
- Probable Prime Number
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