MHB Is $p^3+4$ Prime if Both $p$ and $p^2+8$ are Prime Numbers?

[Albert1]

both $p$ and $p^2+8$ are prime numbers

prove :$p^3+4 $ is also prime

 

[kaliprasad]

both $p$ and $p^2+8$ are prime numbers

prove :$p^3+4 $ is also prime

Spoiler

proof:

p is either 2 or 3 or of the form $(6n\pm1)$

if p is of the form $(6n\pm1)$

then $p^2 + 8 = 36n^2\pm12n + 9 = 3 (12n^2 \pm4n + 3)$ divisible by 3 and the value is greater than 3 so not a prime if p = 2 then $p^2 + 8=12$ 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 given criteria and $p^3+ 4 = 31$ is a prime as well

 

[Albert1]

both $p$ and $p^2+8$ are prime numbers
prove :$p^3+4 $ is also prime

Spoiler

$p\neq 2$ for $2^2+8=12$ is not a prime
let $p=3k=3 (\,\,here \,\,k=1)---(1)$
then $3^2+8=17,$ and $3^3+4=31$ both are prime
if $p=3k+1(k\in N)---(2)$ then $p^2+8=9k^2+6k+9$ not prime
if $p=3k+2(k\in N)---(3)$ then $p^2+8=9k^2+12k+12$ not prime
so the only possible solution is $p=3$ and the proof is done

 

[Greg]

Spoiler

I claim that$$p^2+8\equiv0\mod{3}\quad\forall\,p\ne3$$Consider$$p^2\equiv1\mod{n}$$$$p^2-1\equiv0\mod{n}$$$$(p-1)(p+1)\equiv0\mod{n}$$Now,$$p-1,p,p+1$$are three consecutive integers, hence 3 divides $p-1$ or 3 divides $p+1$ (if $p\ne3$), so we have$$p^2\equiv1\mod{3}\quad\forall\,p\ne3$$hence$$p^2+8\equiv0\mod{3}\quad\forall\,p\ne3$$Also,$$3^2+8=17$$is prime and$$3^3+4=31$$is prime, as required.$$\text{ }$$