A question in quora asked to find real solution(s) to
$x\sqrt{y}+y\sqrt{x} = 6, x+y = 5. $
I showed that the solution with $x \le y$ is $x = 1, y = 4 $.
This naturally brings up the question for which positive integers $a$ and $b$ do the equations
$x\sqrt{y}+y\sqrt{x} = a, x+y = b $
have integer $x$ and $y$ as solutions.
It is clear that for any integers $1 \le p \le q$, $a = pq(p+q), b = p^2+q^2$ has the solution (with $x \le y$) of $x = p^2, y = q^2$.
The original question has $p=1, q=2$.
My question is: are there any other integral $a$ and $b$ for which the equations have integral solutions?
Note: If we just try to solve for $x$, this happens (with the help of Wolfy):
$\begin{array}\\ y &= b-x\\ a &=x\sqrt{y}+y\sqrt{x}\\ &=x\sqrt{b-x}+(b-x)\sqrt{x}\\ a-x\sqrt{b-x} &=(b-x)\sqrt{x}\\ a^2-2ax\sqrt{b-x}+x^2(b-x) &=x(b-x)^2\\ &=x(b^2-2bx+x^2)\\ 2ax\sqrt{b-x} &=x^3-2bx^2+b^2x -a^2-(bx^2-x^3)\\ &=2x^3-3bx^2+b^2x-a^2\\ 4a^2x^2(b-x) &=(2x^3-3bx^2+b^2x-a^2)^2\\ 0 &=a^4 - 2 a^2 b^2 x + 2 a^2 b x^2 + b^4 x^2 - 6 b^3 x^3 + 13 b^2 x^4 - 12 b x^5 + 4 x^6\\ \text{with real roots}\\ x &= \dfrac12 \left(b \pm \sqrt{b^2 - 4 \left(-\dfrac{r}{12} + \dfrac{(24 a^2 b - b^4)}{12 r} + \dfrac{b^2}{12}\right)}\right)\\ \text{where}\\ r &=\left(-216 a^4 + 36 a^2 b^3 + 24 \sqrt{3} \sqrt{27 a^8 - a^6 b^3} - b^6\right)^{1/3}\\ \end{array} $
I don't know how much help this is.
If $$x\sqrt{y}+y\sqrt{x} = a,\tag{1}$$ with $x,y,a\in \mathbb{N},$ then $$x^2y+2xy\sqrt{xy}+xy^2=a^2,$$ so that $\sqrt{xy}$ is rational, and $xy$ is a perfect square. Let $g=\gcd(x,y),$ so that $x=gm, y=gn$ where $m$ and $n$ are co-prime integers. We have that $\sqrt{nm}$ is rational, so that $nm$ is a perfect square, and since $\gcd(n,m)=1$ $m$ and $n$ are each perfect squares, say $x=gr^2,\ y=gs^2,\ r,s\in\mathbb{N}$.
Substituting these values into $(1)$ we get that $\sqrt{g}$ is rational, so that $g$ is a perfect square, and finally, that $x$ and $y$ are perfect squares.
In short, you have found all the possibilities.