Two Related Primes

If $p$ is a prime it can either leave remainder of 1 or 2 when divided by 3, (or 0 in case the number is 3 itself). Now if it leaves remainder 1 then $(p^{2} + 8) | 3 \equiv 0$ and if it leaves remainder 2, again $(p^{2} + 8) | 3 \equiv 0$ . (Check?)

Hence the only value possible for $p$ is $\boxed{3}$ . Answer

$p\equiv\pm1 mod(6) \Rightarrow\ p^2 +8\equiv3mod(6)$ This means that for both ${p}$ and $p^{2}$ + 8 to be prime, then: $p\neq\pm1 mod(6)$ $and \ p \ has \ to \ be \ odd$ $\therefore\ p = 3 \ and \ p^2 + 8 = 17 \ is \ the \ only \ solution$

$\forall p \in \mathbb{P}, \; p \geq 5, \; p \equiv \pm 1 \pmod{6}$

$\therefore$ if $\left\{p, p^2+8 \right\} \subset \mathbb{P}$

Then $p^2 + 8 \equiv \pm 1 \pmod{6}$

$\Rightarrow p^2 \equiv -9$ or $-7 \pmod{6} \equiv -3$ or $-1\pmod{6}$

However, $p \equiv \pm 1 \pmod{6} \Rightarrow p^2 \equiv 1 \pmod{6}$

This is a contradiction and thus we see that there are no such primes for $p \geq 5$ , so we need only check $p = 2$ and $p =3$

$p =2 \Rightarrow 2| \left(p^2 +8 \right) \Rightarrow \left(p^2+8\right) \notin \mathbb{P}$

$p = 3 \Rightarrow p^2 + 8 = 17 \in \mathbb{P}$

Therefore, the sum of all values of $p$ that satisfy the equations is $3$

$p^2 + 8 = p^2 - 1 + 9 = (p - 1)(p + 1) + 9$ Considering this equation, if $p$ is not divisible by $3$ , then either $p - 1$ or $p + 1$ will be, since they are three consecutive numbers, and $p^2 + 8$ will not be prime. If $p = 3$ , then $p^2 + 8 = 17$ , so $3$ is the only solution.

Let's consider

1. p = 3m-1 for m = any integer, then $p^{2}+8$ = $9m^{2}-6m+9$ = $3(3m^{2}-2m+3)$ which is dividable by 3
2. p = 3m-2 for m = any integer, then $p^{2}+8$ = $9m^{2}-12m+12$ = $3(3m^{2}-4m+4 )$ which is also dividable by 3
3. p = 3m for m = any integer, then $p^{2}+8$ = $9m^{2} + 8$ which might be prime but p = 3m is prime only when m = 1

Sp, p = 3 only solution.

Using Fermat's little theorem: (note that this holds true when p is relatively prime to 3)

$p^2$ = 1 ( mod 3 ), therefore $p^2$ + 8 = 9 = 0 ( mod 3 ).

This tells us that any number that is relatively prime to 3 (which includes every prime other than 3 itself) is divisible by 3.

This means the only prime we need to test is 3. $3^2$ + 8 = 17, which is prime. This tells us that the only number that works is 3 itself

Every prime $p\neq3$ is $\pm1\ (\text{mod}\ 3)$ . ( CRUX MOVE )

This implies $p^2\equiv1\ (\text{mod}\ 3)$ and so $p^2+8\equiv9\equiv0\ (\text{mod}\ 3)$ The only remaining case is $p=3$ , which satisfies the condition. Thus, the answer is $\boxed3$ .

Apart from number 3, that is the correct answer, for any other possible prime with the expression p^2 + 8, that result should never be equal to a multiple of 3.

Every prime number, apart from number 3 itself, is either of the form 3x + 1 or the form 3x - 1. If we substitute our original "p" by these two expressions, we get that new formed primes should be of the value 9x^2 + 6x + 9 or 9x^2 - 6x + 9, which is always going to be a multiple of 3, and therefore not prime.