Suppose that $p$ and $p^2 + 2$ are primes show that $p^3 + 2$ is prime

Question

Can someone help me out with this? I've been working on it for quite a long time but I'm not sure if I'm even getting anywhere.

Answer 1

This answer has the same basic reasoning as all the other answers, but from a slightly different angle. You know that the only even prime is $2$, and all other primes are odd.

$p\ne 2$ since $2^2+2=6$ which is not prime, so $p$ must be odd.

Every odd prime except $3$ has the form $6k\pm 1$, and $(6k\pm 1)^2$ will have the form $6j+1$.

$6j+1+2=6j+3$ which is divisible by $3$ and therefore not prime. So primes of the form $6k\pm 1$ are excluded, and the only option left is $p=3$, which in fact satisfies the requirements $p^2+2=11'\space p^3+2=29$

Answer 2

Hint: There are very few primes $p$ for which $p^2+2$ is also prime.

Try it for a few different primes and see if you can spot a pattern in what way the $p^2+2$ all fail to be prime.

Answer 3

If $p\ne 3$ then $p^2+2\ge 6$ is a multiple of $3$ and not prime; so $p=3$, and $p^2+2=11,\,p^3+2=29$ are both prime.

Answer 4

They key idea is that $p$ and $p^2+2$ are never both prime, unless . . .

Unless $p=3$.

Clearly we can't have $p=2$.

Suppose $p > 3$.

Note that for any integer $x$, one of $x-1,x,x+1$ is a multiple of $3$.

Since $p$ is not a multiple of $3$, it follows that $p^2+2=(p-1)(p+1)+3$ is a multiple of $3$.

But $p^2+2 > 3$, hence, since it's a multiple of $3$, it can't be prime.

So the only remaining case is $p=3$, which works, since $3^2+2=11$ and $3^3+2=29$.