Are there integer solutions such that the sum of the roots of x and y equal the root of a prime p? How could one prove whether or not integer pairs (x, y) exist such that $\sqrt{x} + \sqrt{y} = \sqrt{p}$, where p is prime? `

Using elementary methods.
We'd have
$$x + 2\sqrt{xy} + y = p$$
and hence
\begin{align}
x + y - p = -2\sqrt{xy}
&\implies
x^2+y^2+p^2+2xy-2xp-2yp = 4xy
\\&\iff
(x^2-2xy+y^2) + p^2= 2p(x+y)
\\&\iff
(x-y)^2 +p^2 = 2p(x+y)\tag{1}
\\&\iff
(x-y)^2 +2p(x-y) +p^2 = 2p(x+y) + 2p(x-y)\tag{2}
\\&\iff
(x-y+p)^2 = 4px.
\end{align}
This implies $x-y+p = \pm 2\sqrt{px}$, so $x = pa^2$ for some integer $a$.
At $(2)$, we added $2p(x-y)$ to both sides of equation $(1)$.
If we instead had subtracted $2p(x-y)$, we'd get
\begin{align}
x + y - p = -2\sqrt{xy}
&\implies
(x-y)^2 -2p(x-y) +p^2 = 2p(x+y) - 2p(x-y)\tag{3}
\\&\iff
(x-y-p)^2 = 4py.
\end{align}
Like before, we obtain that $x-y-p = \pm 2\sqrt{py}$, and hence $y = pb^2$ for some integer $b$.
The initial equation can then be rewritten:
$$|a|\sqrt p + |b|\sqrt p = \sqrt p \iff |a| + |b| = 1.$$
This can happen only when $|a| = 1$ and $b=0$ or vice versa, that is, either $x = p$ and $y=0$ or $x=0$ and $y=p$.
