<s>[INST] 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? `
 [/INST] A slightly less elementary approach. Consider the polynomial whose roots are the pairwise sums of the roots of the minimal polynomials of $\sqrt x$ and $\sqrt y$: $$(z - \sqrt x - \sqrt y)(z - \sqrt x + \sqrt y)(z + \sqrt x - \sqrt y)(z + \sqrt x + \sqrt y) \\
= ((z - \sqrt x)^2 - y) ((z + \sqrt x)^2 - y) \\
= (z^2 + x - y - 2z\sqrt x) (z^2 + x - y + 2z\sqrt x) \\
= (z^2 + x - y)^2 - 4z^2x \\
= z^4 - 2(x+y) z^2 + (x-y)^2$$
If $\sqrt p$ is a root then its minimal polynomial $z^2 - p$ is a factor. Since we have two monic polynomials, that implies that $p$ is a factor of $(x-y)^2$. The tail of Elliot's argument wraps it up.
We could instead have gone on to do a long division, finding that the remainder is $$p^2 - 2(x+y)p + (x-y) ^2 =0$$ as in Fimpellizieri's answer. 
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