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April 11th, 2017, 02:50 AM  #1 
Newbie Joined: Apr 2017 From: poland Posts: 1 Thanks: 0  Pythagorean prime in every prime twin
Hi, I have a thread to understand, but I have some problem with it. I must prove that in every prime twin there is one and only one Pythagorean prime. I found something about it on 5 page of: www.fq.math.ca/Scanned/242/sternheimer.pdf Can anyone explain it a bit easier? Thank you a lot. Last edited by skipjack; April 12th, 2017 at 04:44 AM. 
April 12th, 2017, 02:21 AM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
A Pythagorean Prime is of the form $4n + 1$ so the best you can do is to have $2$ consecutive Pythagorean Primes differ by $4$ , like $13$ and $17$ Twin primes differ by $2$ exactly so it is impossible to have $2$ Pythagorean Primes in any set of twin primes $(p_1 , p_2)$ Last edited by agentredlum; April 12th, 2017 at 02:39 AM. Reason: fixed error 
April 12th, 2017, 02:35 AM  #3 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
Any prime greater than $2$ must be of the form $4n + 1$ or $4n 1$ Consider the arbitrary twin prime ordered pair $(p_1, p_2)$ with $p_1 < p_2$ and $2$ cases Case 1 If $p_1$ is of the form $4n + 1$ then $p_2$ must be of the form $4n + 3$ so only $1$ Pythagorean Prime in this case Case 2 If $p_1$ is of the form $4n 1$ then $p_2$ is of the form $4n + 1$ so only $1$ Pythagorean Prime in this case This exhausts all possible cases for twin primes given your conditions 

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