-1
$\begingroup$

Let $p$ be a prime number such that $p^2+12$ has exactly $5$ divisors. What is the maximum value of $p$ ?

I came across this question in a Math Olympiad Competition and had no idea how to solve it

$\endgroup$

closed as off-topic by TheSimpliFire, Lord_Farin, user21820, Did, RRL Dec 25 '18 at 14:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Lord_Farin, user21820, Did, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ A number with an odd number of divisors must be square. $\endgroup$ – Lord Shark the Unknown Dec 25 '18 at 6:02
8
$\begingroup$

Notice that a number that has 5 divisors must be in the form of $q^{4}$ for some prime $q$. So we get:

$q^{4}=p^{2}+12 \implies (q^{2}-p)(q^{2}+p)=12$

Then do some casework on it.

$\endgroup$
  • $\begingroup$ The answer I have got is 2(maximum value of P). I think I am right ? $\endgroup$ – Mohammad Mizanur Rahaman Dec 25 '18 at 6:13
  • 1
    $\begingroup$ @MohammadMizanurRahaman: yes, $p=q=2$ $\endgroup$ – Ross Millikan Dec 25 '18 at 6:24
0
$\begingroup$

Using @LordSharktheUnknown's hint that a number with an odd number of divisors must be square

we are then looking for cases where $p^2+12=a^2$ i.e. $a^2-p^2 = 12$

Since $7^2-6^2=13$, this tells you $a^2-p^2 \gt 12$ for $a \ge 7$ and $p \lt a$

so here we must have $a \lt 7$ and thus prime $p \in \{2,3,5\}$. Considering these:

  • $2^2+12=16=4^2$ so $p=2$ is indeed a possibility
  • $3^2+12=21$, not a square, so $p=3$ is not a possibility
  • $5^2+12=37$, not a square, so $p=5$ is not a possibility
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.