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.$$\text{ }$$

 

[CellCoree]

I understand the importance of evidence and rigorous proof in any scientific claim. In this case, we are trying to prove that $p^3+4$ is prime, given that both $p$ and $p^2+8$ are prime numbers.

First, let's start by defining what it means for a number to be prime. A prime number is a positive integer that is only divisible by 1 and itself. This means that a prime number cannot be divided by any other number to give a whole number result.

Now, let's look at the expression $p^3+4$. We can rewrite this as $(p^2)^2+4$. This suggests that $p^3+4$ is in the form of a sum of squares, which can be factored using the difference of squares formula: $a^2-b^2 = (a-b)(a+b)$.

Applying this formula to $p^3+4$, we get $(p^2+2)(p^2-2)$. Now, we know that $p^2+8$ is prime, so $p^2+2$ must be a factor of $p^2+8$. This means that $p^2+2$ cannot be divided by any other number to give a whole number result, making it a prime number.

Similarly, $p$ is also a prime number, so it cannot be divided by any other number to give a whole number result. This means that $p^2-2$ cannot be divided by any other number to give a whole number result, making it a prime number as well.

Putting this all together, we have shown that $p^3+4$ can be factored into two prime numbers, $p^2+2$ and $p^2-2$. Since these are the only factors of $p^3+4$, we can conclude that $p^3+4$ is a prime number.

In conclusion, we have proven that $p^3+4$ is prime, given that both $p$ and $p^2+8$ are prime numbers. This proof relies on the definition of prime numbers and the difference of squares formula. Therefore, we can confidently say that $p^3+4$ is indeed a prime number.