proof:
p is either 2 or 3 or of the form (6n+/-1)
if p is of the form (6n+/-1)
then p^2 + 8 = 36n^2-12n + 9 = 3 ( 12n^2 - 4n + 3) divisible by 3 and ( 12n^2 - 4n + 3) > 1
so not a prime
if p = 2 then p^2 + 8 is even and not a prime
so only p =3 is left and p^2+ 8 = 17 is a prime and p =3 is the only number satisfies the criteria and p^3+ 4 = 31 is a prime as well
2 comments:
Can you teach me why "p is either 2 or 3 or of the form (6n+/-1)" ? Thanks.
2 and 3 primes
now the number is of the form
6n,
6n+/-2
6n+3
6n+/-1
first 3 forms( 6n= 2* 3n, so on) cannot be prime so prime have to be of the form 6n+/-1
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