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 c≥1
Any prime >3 can be written as 6a±1 where a≥1
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+(3c−1)=3(3b2±2b+c).
Then , p2+(3c−1) is divisible by 3
and p2+(3c−1)>3 if p>3 and c≥1,hence not prime.
The necessary condition for p2+(3c−1) to be prime is 3∣p
⟹ if p2+(3c−1) is prime, 3∣p.
If p needs to be prime, p=3, here c=3
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