For any Pythagorean triple, the number of primes under a + b + c must


by goldust
Tags: number, primes, pythagorean, triple
goldust
goldust is offline
#1
Oct29-13, 03:34 PM
P: 85
be no more than c? In fact, only for the first triple does equality hold. Upon examining some of the triples, I noticed this must be true. However, I'm having a hard time explaining why. Is there a good explanation for this? Many thanks!
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Office_Shredder
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#2
Oct29-13, 04:25 PM
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Since c is larger than a or b, you're basically saying the number of primes smaller than 3c is less than c... for c sufficiently large this is because the number of primes smaller than n is log(n). So the only worry would be that for c small you could have a counterexample and it just turns out there isn't one I guess. There might be a more solid reason but I would guess this is probably all that's happening.
willem2
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#3
Oct30-13, 03:12 AM
P: 1,351
Quote Quote by Office_Shredder View Post
Since c is larger than a or b, you're basically saying the number of primes smaller than 3c is less than c... for c sufficiently large this is because the number of primes smaller than n is log(n).
the number of primes smaller than n is approximately n/log(n), or more precisely:

lim n→∞ (pi(n) log (n)) / n = 1

where pi(n) is the number of primes smaller than n. (prime number theorem)

You don't really need the prime number theorem here. If you only consider division by 2,3 and 5 it's easy to see that pi(n)< (8/30)n + 8 (because n mod 30 must be in {1,7,11,13,17,19,23,29})


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