# Prime number and Divisors [closed]

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

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

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

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.

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

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