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Determine all prime numbers p such that p^2+2 is also a prime number.

I have only found 3, and I think that's the only number but I have trouble writing proof for this. But I know that if p is a prime number other than 3 then p^2+2 will be divisible with 3.

amWhy
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Negative Creep
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2 Answers2

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Suppose $p$ is prime and different from $3$. Write $p^2+2$ as $$p^2+2=(p^2-1)+3=(p-1)(p+1)+3.$$ Since $3\nmid p$, $3$ has to divide either $p-1$ or $p+1$. We conclude that $3$ divides $(p-1)(p+1)+3=p^2+2$, and since $p^2+2\ne 3$, it follows that $p^2+2$ is not prime.

Lios
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  • I don't quite follow how we can conclude that 3 divides (p-1)(p+1)+-... when 3 is different from 3 – Negative Creep Nov 07 '20 at 10:02
  • The fact is that if $a,b,c$ are three consecutive natural numbers, then $3$ has to divide exactly one of them. Can you see why? Therefore, since $(p-1),p,(p+1)$ are three consecutive numbers and $3$ does not divide $p$, we deduce that $3$ divides either $p-1$ or $p+1$. – Lios Nov 07 '20 at 10:13
  • Then if $3$ divides one between $p-1$ and $p+1$, it has to divide also their product, and since $3$ clearly divides $3$, it follows that $3$ divides the sum $(p-1)(p+1)+3$. – Lios Nov 07 '20 at 10:14
  • How can I prove that all other primes in p^2 + 2 other than 3 är divisible by 3? I need to prove that first I guess? – Negative Creep Nov 07 '20 at 11:11
  • Do you agree that if $p^2+2$ is divisible by $3$, and $p^2+2\ne 3$, then $p^2+3$ is not prime? Hence the only thing we need to prove is that $p^2+2$ is divisible by $3$, and that is axactly the content of my answer. (how can a prime different from $3$ be divisible by $3$?) – Lios Nov 07 '20 at 11:19
  • No, I don't follow, I just don't know how to write this down. I can't just start off supposing p is prime different from 3, where did 3 come from or why am I even saying p is different than 3. I understand that if p is a prime number other than 3 then p^2+2 will be divisible with 3. But it's weird to start a proof like that just supposing that p is a prime different from 3. I just don't know how to prove my conclusion that p^2 + 2 is always divisible by 3 if p is not 3. – Negative Creep Nov 07 '20 at 12:36
  • Or shall I just start off claiming that if p is a prime number other than 3 then p^2+2 will be divisible with 3. Or is it enough to show some examples? – Negative Creep Nov 07 '20 at 13:00
  • It is actually simpler: you just say that if $p=3$, then $p^2+2=11$ is a prime, while if $p\ne 3$, then $3$ divides $p^2+2$ and $3\ne p^2+2$ (like I said in the answer) therefore $p^2+2$ is not prime. It is a perfectly fine proof. – Lios Nov 07 '20 at 13:17
  • Yes, I understand it now, I only have one question, is 3≠^2+2 because you want to show that 1 is not a prime, or what do you mean by it? – Negative Creep Nov 07 '20 at 13:50
  • I just mean that to show that a number $x$ is not prime, it is not enough to find a prime number $p$ which divides $x$, since, for example, $5$ is a prime, $5$ divides $5$, but this of course does not mean that $5$ is not prime. – Lios Nov 07 '20 at 13:55
  • I understand your proof I just don't follow your summary, I understand that if p≠3 then 3 divides ^2+2 but then you said "and 3≠ p^2+2", I'm being stupid but I don't get this expression. – Negative Creep Nov 07 '20 at 14:18
  • Let us continue this discussion in chat. – Lios Nov 07 '20 at 14:23
1

Primes larger than $3$ cannot be divisible by either $2$ or $3$. Numbers with this property, which include all primes other than $2$ or $3$, have the form $6k\pm 1$.

$p^2+2=(6k\pm 1)^2+2=6k^2\pm 12k+3=3(2k^2\pm 4k+1)$, showing that $3\mid (p^2+2)$ when $p\ne 2,3$.

Simple arithmetic checking shows that only $p=3$ has the property that $p^2+2=11$ yields a prime.

Keith Backman
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