Obtained from here.
"Problem 1.
(a) Let the positive integers $p,\;q$ be prime numbers and let $a$ be a positive
integer. If $a$ divides the product $p\cdot q$ , and it holds that $a>p$ and $a>q$ ,
prove that $a=pq$.
(b) Determine all pair $(p,q)$ of prime numbers such that $p^2+3pq+q^2$
eqals a perfect square."
ANSWER CiP
(b) $(3,\;7)$ and $(7,\;3)$
Solution CiP
(a) $a\mid p\cdot q\;\Rightarrow\;\;p\cdot q=a\cdot b$ for a certain $b\in \mathbb{N}$. Noting that $p$ divides the product $a\cdot b$ then, since it is prime, it follows
$$p\mid a\;\;\;or\;\;\;p\mid b.$$
If $p\mid a$, then $a=p\cdot c$ for a certain $c\in \mathbb{N}$. Then, from
$p\cdot q= (p\cdot c)\cdot b$ we obtain $q=c\cdot b$. But $q$ is also prime so we can have $c=1$ (when $q=b,\;p=a$ which contradicts $a>p$) or $b=1$ , in which case $q=c$ and $a=pq$.
If $p\mid b$, then $b=p\cdot d$ for a certain $d\in\mathbb{N}$. Then, from $p\cdot q=a(p\cdot d)$ we obtain $q=a\cdot d$, but $q$ being prime and $a>1$ it follow $d=1,\;q=a$ wich contradicts $a>q$.
With these, (a) is proven.
(b) If $k\in \mathbb{N}$ is such that $p^2+3pq+q^2=k^2$ then we get
$$pq=k^2-p^2-2pq-q^2=k^2-(p+q)^2=(k-p-q)\cdot (k+p+q)$$
From the above it can be seen that the number $a:=k+p+q>p,\;q$ divides the product $pq$. According to (a) we must have $a=pq$ that is, equivalent to
$k+p+q=pq\;\Rightarrow\;k=pq-p-q\;\;\Rightarrow\;\;p^2+3pq+q^2=(pq-p-q)^2\;\:\Leftrightarrow$
$\Leftrightarrow\;pq=p^2q^2-2p^2q-2pq^2\;\;\Leftrightarrow\;\;1=pq-2p-2q\;\Leftrightarrow\;5=pq-2p-2q+4\;\Leftrightarrow$
$$\Leftrightarrow\;\;\;5=(p-2)(q-2).$$
Then it follows that $p-2=1$ or $p-2=5$. We get the answer.
Verification: $3^2+3\cdot 3\cdot 7+7^2=9+63+49=121=11^2$.
$\blacksquare$
