Thursday, 28 August 2014

elementary number theory - Prime p with p2+8 prime



I need to prove that there is only one p prime number such that p2+8 is prime and find that prime.



Anyway, I just guessed and the answer is 3 but how do I prove that?


Answer




Any number can be written as 6c,6c±1,6c±2=2(3c±1),6c+3=3(2c+1)



Clearly, 6c,6c±2,6c+3 can not be prime for c1



Any prime >3 can be written as 6a±1 where a1



So, p2+8=(6a±1)2+8=3(12a2±4a+3).



Then , p2+8>3 is divisible by 3,hence is not prime.




So, the only prime is 3.






Any number(p) not divisible by 3, can be written as 3b±1



Now, (3b±1)2+(3c1)=3(3b2±2b+c).



Then , p2+(3c1) is divisible by 3




and p2+(3c1)>3 if p>3 and c1,hence not prime.



The necessary condition for p2+(3c1) to be prime is 3p



if p2+(3c1) is prime, 3p.



If p needs to be prime, p=3, here c=3


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