My solution: Firts, we require that $ \frac{p}{n} $ is an integr, so either $ n=1$ or $p$. In any case, $ p + 1 = a^2 $ for some natural $a$. But then $ p = (a-1)(a+1)$ so $ a - 1 = 1 $. Hence $ a = 2 $ so the only solution is $ n,p = 1,3$ and $ 3,3$.

I think the question should be: fprosk wrote: Find all pairs of natural numbers n and prime numbers p such that $\sqrt{n+\frac{p}{n}}$ is a rational number. Because the original one is so easy for a TST!

Let $\sqrt{n+\frac{p}{n}}=\frac{a}{b}$ or $\frac{n^2+p}{n}=\frac{a^2}{b^2}$ Let $(n^2+p,n)=d$ $\Leftrightarrow d \mid n$ and $d\mid n^2+p\Rightarrow d\mid p$ hence $d=1$ or $d=p$ If $d=1$ then $n=b^2$ and $n^2+p=a^2\Leftrightarrow p=(a-n)(a+n)\Rightarrow p=2n+1$ Therefore $(n,p)=(b^2,2b^2+1)$ If $d=p$ then $n^2+p=pa^2\Leftrightarrow n^2=p(a^2-1)$ yielding $p$ is a perfect square (a contradiction) or $p=a^2-1=(a-1)(a+1)$ hence $n=p=3$ In conclusion, $(n,p)=(k^2,2k^2+1);(3,3)$

$ \sqrt{n^3+pn}$ is integer $n^3+pn=a^2$ $d=(n,a)$ $d^2 |a^2-n^3=pn$ Case 1: $d|p \to d=1,p$ $d=1$ and $n |a^2 \to n=1$ $p+1=a^2 \to p=3,a=2$ $d=p$ $n=pn_1,a=pa_1$ $pn_1^3+n_1=a_1^2$ $n_1 |a_1^2$ and $(a_1,n_1)=1 \to n_1=1$ $n=p$ $p^3+p^2=p^2a_1^2$ $p=a_1^2-1 \to p=3,a_1=2,n=3,a=6$ Case 2: $(d,p)=1, d \neq 1$ $d^2 |pn \to d^2| n$ $n=d^2n_1,a=da_1, (n_1,a_1)=1$ $d^6 n_1^3+pd^2n_1=d^2a_1^2$ $(d^4n_1^2+p)n_1=a_1^2$ $n_1 |a_1^2 \to n_1=1 \to n=d^2$ $d^4+p=a_1^2$ $p=(a_1-d^2)(a_1+d^2)$ $a_1=d^2+1,p=2d^2+1, a=d(d^2+1)$ If $d \neq 0$ mod $3$ , than $2d^2+1 \equiv 0$ (mod 3) , so $p$ is not prime (except $d=1$) $d=3k$ $n=9k^2,p=18k^2+1$ I think there are many such $k$ that $18k^2+1$ is prime. Answer: $(1,3),(3,3),(9k^2,18k^2+1)$ where $18k^2+1 $ is prime.

$n\mid p,\;n\in\{1,p\}$ since $\frac{n^2+p}{n} = a^2,\; a\in\mathbb{N}^+,\;p\in\mathbb{P}$. So $p+1=\frac{n^2+p}{n}=a^2,\;p = (a-1)(a+1)\in\mathbb{P}\implies p = 3$ Consequently, $(1,3),(3,3)$ are the complete solutions $(n,p)$.