2. find all $p\in \mathbb{P} $ that $P^{2}+11$ has only 6 positive factor
please see what I do wrong
$\because P^{2}+11$ has 6 positive factor $\therefore P^{2}+11=mn^{2},r^{5}$ that$m,n,r\in \mathbb{P} $
Case 1
$P^{2}+11=mn^{2}$
$11=m(n^{2}-\frac{p^{2}}{m})$
$\because 11\in \mathbb{P} ,m\in \mathbb{P} \therefore m=11$
$1=n^{2}- \frac{p^{2}}{11}$
$P^{2}=11(n^{2}-1)$
$\therefore P=11,n= \sqrt{12}$ --->n not true so no answer for this case
Case 2
$P^{2}+11=r^{5}$
$11=r(r^{4}-\frac{P^{2}}{r})$
$\because 11\in \mathbb{P} ,r\in \mathbb{P} \therefore r=11$
$11=11^{4}-P^{2}$
$P^{2}=11^{5}-11$
$\sqrt{11^{5}-11} \not\in \mathbb{P} \therefore$ no answer for this case
from case1,2 $\therefore$ no anwser
BUTTT!! if i give $P=3--> P^{2}+11=20=2^{2}\times 5$ there is an anwser?
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