Show that $p^3+4$ is prime

Question

If $p$ and $p^2 +8$ are both prime number, prove that $p^3 +4$ is also prime.

I was thinking using two cases for even and odd. So $p$ is even thus there is no prime, the statement does not hold. So, $p$ is odd, $p$ has to be of the form of $2k+1$.

let $p=2k+1$ which is prime

$(2k+1)^2 + 8 = 4k^2+4k+1+8 = 2(2k^2+2k+4)+1$ which is prime

so $(2k+1)^3 +4 = 8k^3+12k^2+6k+1 +4= 2(4k^3+6k^2+3k+2)+1$ which is also prime.

is that all that i need to show

Note: 2 is even but also prime. However, for p=2, note that p^2+8=12, which isn't prime. Thus, p=2 isn't even a candidate meeting the hypotheses. You can make an observation of this in your proof, and then just deal with the case of p odd. — Sinister Cutlass, Sep 28 '15 at 03:08
@SinisterCutlass In reality, you don't need to take $2$ as a special case at all. — Thomas Andrews, Sep 28 '15 at 03:11
Yeah i see that, that is one of the reason why i post it. I did feel like the proof did not complete — user146269, Sep 28 '15 at 03:12
This seems to be a duplicate of Why does $p^2+8$ prime imply $p^3+4$ prime. See also: Prime $p$ with $p^2+8$ prime — Martin Sleziak, Sep 28 '15 at 15:55

score 8 · Answer 1 · answered Sep 28 '15 at 03:09

8

No, as commenters have points out, all you've shown is that $p^3+4$ is odd.

Hint: What must $p$ be, modulo $3$?

answered Sep 28 '15 at 03:09

Thomas Andrews

177,126

what do you mean by what must p be modulo 3, and why modulo – user146269 Sep 28 '15 at 03:13
Every natural number is of the form $3k,$, $3k+1$, or $3k+2$, which are called in number theory 0, 1 and 2, modulo $3$, respectively. – Thomas Andrews Sep 28 '15 at 03:15
1

Very nice solution. – Ben Sheller Sep 28 '15 at 03:17

score 0 · Answer 2 · edited Oct 31 '18 at 21:30

0

Hint: if you don't know the theory of quadratic residue, you can note that every prime $p>3$ can be written in the form $6n\pm1$. It has remainder $\pm1$ when divided by $3$. Note therefore that for $p>3$, $p^2+8$ is divisible by $3$. Indeed $$p^2+8=(6n\pm1)^2+8\equiv (\pm1)^2\!+8\equiv 0\pmod{\!3}$$

edited Oct 31 '18 at 21:30

Bill Dubuque

272,048

answered Sep 28 '15 at 15:22

Domenico Vuono

2,672

$5$ and $11$ are not of the form $6n+1$ :-) – Quang Hoang Sep 28 '15 at 15:34
But 6n\pm 1 yes – Domenico Vuono Sep 28 '15 at 15:36
But why $6n\pm1$ when you can do $3n\pm1$? – Quang Hoang Sep 28 '15 at 15:38
$3n\pm 1 $ is even, it doesn't rappresent a prime number – Domenico Vuono Sep 28 '15 at 15:40
But does one need that? Did you read Thomas' answer? – Quang Hoang Sep 28 '15 at 15:47

Show that $p^3+4$ is prime

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