7.8. The natural number $n$ and the prime numbers $p$ and $q$ satisfy the relation $\frac{3}{\sqrt{n}}=\frac{2}{\sqrt{p}}+\frac{1}{\sqrt{q}}$. Determine all possible values of the expression $E=p+q-2 n$.
## Solution:

From $\frac{3}{\sqrt{n}}=\frac{2}{\sqrt{p}}+\frac{1}{\sqrt{q}}$, we get $\frac{3}{\sqrt{n}}=\frac{2 \sqrt{q}+\sqrt{p}}{\sqrt{p q}}, 3 \sqrt{p q}=\sqrt{n}(2 \sqrt{q}+\sqrt{p})$. Squaring both sides of the last equality, we obtain $9 p q=n(4 q+p+4 \sqrt{p q})$. Since $9 p q \in \mathbb{N}$, it follows that $\sqrt{p q} \in \mathbb{N}$, which implies $p \cdot q$ - a perfect square. Since $p$ and $q$ are prime numbers, and $p \cdot q$ - a perfect square, it follows that $p=q$. For $p=q$, from $9 p q=n(4 q+p+4 \sqrt{p q})$ we get $9 p^{2}=9 p n$, which implies $p=n$. Thus, $n=p=q$, and the expression $E=p+q-2 n=0$.

## Answer: 0