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

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The discussion centers on the relationship between Pythagorean triples and the distribution of prime numbers. It establishes that for any Pythagorean triple (a, b, c), the number of primes less than a + b + c is no more than c, with equality holding only for the first triple. The reasoning involves the prime number theorem, which states that the number of primes less than n is approximately n/log(n). The conclusion emphasizes that for sufficiently large c, the number of primes below 3c is less than c, reinforcing the established relationship.

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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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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.
 
Office_Shredder said:
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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