Let's have the number $5^{2^{n-1}}$ where $n$ any non-zero natural number. I conjecturally say that between the following two bounds we will always obtain $n$ primes of the form $4x+1$. $[(5^{2^{n-1}})^{1/n}]e^{1/n}<...>[(5^{2^{n-1}})^{1/n}]e^{-1/n}$
Prime of the form $4x+1$ within two bounds
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0"the two other well known methods"? – 2011-11-23
2 Answers
The interval in question is essentially of the form$\bigg[T\bigg(1-\frac c{\log\log T}\bigg),T\bigg(1+\frac c{\log\log T}\bigg)\bigg]$ for $T=5^{2^{n-1}/n}$. Chebyshev-type bounds will not be strong enough to establish that there are primes in this interval. However, the prime number theorem for arithmetic progressions is strong enough to show that there are asymptotically $cT/(\log T \log\log T)$ primes in that interval that are 1 (mod 4). Therefore your conjecture is true when $n$ is sufficiently large, and could in principle be confirmed for all $n$ by a finite calculation.
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0Sorry, unless you calculate by brute force in any given instance, your question does not lend itself to exact answers. – 2011-11-25
Erdos proved the Chebyshev bounds for the prime counting function at a very young age. You can find these proofs in Hardy-Wright. After a short period of time he also proved the analogous result for the counting function of the primes $\pm 1 \pmod 4.$
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0My question was directed to esofos. – 2011-11-23