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How do I prove that $p^2+8$ prime implies that $p^3+4$ is prime? What is the general pattern of thought for these problems?

Jyrki Lahtonen
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Temitope.A
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    Show that the only such prime is $p=3$. This is a 'standard problem' in the literature. – Calvin Lin Jan 03 '13 at 16:55
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    I've now seen this solution, what a trick! ><. To solve this problem one must have tried inserting some values of p, and then make that conjecture. I didn't do it and tried a direct proof. I suppose there isn't in this case. Is it true? – Temitope.A Jan 03 '13 at 17:00
  • Another common trick is that for $p\geq 5$, it is of the form $6k\pm 1$. So if your answers are contained in $2,3$, then using $6k\pm1$ is a worthwhile shot. – Calvin Lin Jan 03 '13 at 17:07
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    Students tend to get sucked into attempting a "direct proof", because so many of the things they are asked to prove are a quick consequence of some recent theorem. In general, one needs to fool around first, to see what's going on. With some experience in number theory, you will know that it is hopeless to prove that $p^3+4$ is prime given that $p^2+8$ is prime, unless $p^2+8$ prime basically can't happen. – André Nicolas Jan 03 '13 at 17:44
  • See also: Prime $p$ with $p^2+8$ prime – Martin Sleziak Sep 28 '15 at 15:57
  • You did not write this in your question, but I guess $p$ is supposed to be prime, too? – Martin Sleziak Oct 03 '15 at 09:30

2 Answers2

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If prime $p\ne 3,p$ can be written as $3k\pm1$ where $k$ is a positive integer.

So, $p^2+8=(3k\pm1)^2+8=3(3k^2\pm2k+3)$ is divisible by $3$ and $>3$ hence composite.

So, we are left with only one prime value of $p,$ i..e, $3$

More generally, $p^2+3m-1$ is composite for any positive integers $m$ and $3\not\mid p$

When we have found $3$ can be only value of prime $p,$ we can have umpteen prime numbers like $p^3+4,p^3+10,p^3-2,\cdots, p^4\pm2,\cdots $

lab bhattacharjee
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The question or (at least the current revision) does not say that $p$ is prime. Although this is very probably the intended meaning. (The letter $p$ is typically used to denote prime numbers. Exactly this exercise with the assumption the $p$ is a prime number appears in many introductory texts in number theory.)

If we take the question in the way it is stated, i.e., we are asking whether $p^2+8$ being prime necessarily implies $p^3+4$ being prime, then the claim is not true.

The arguments in the other answers show us that if want to find a counterexample, it is possible only with $p$ being odd multiple of $3$. And indeed we can find a counterexample: $$\begin{align} 15^2+8&=233\\ 15^3+4&=3379 \end{align} $$ 233 is a prime number, but 3379 is not a prime, in fact we get the factorization $3379=31\cdot109$.

Community
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Martin Sleziak
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