Weird equations

There is an easier solution as shown here.

$\begin{aligned} (\sqrt{x} + \sqrt{y})^3 & = \sqrt{x}^3 + 3\sqrt{x}^2 \sqrt{y} + 3 \sqrt{x} \sqrt{y}^2 + \sqrt{y}^3 \\ & = x\sqrt{x} + 3(x\sqrt{y} + y\sqrt{x} ) + y\sqrt{y} \\ & = 65 + 3(20) = 125 \\ \Rightarrow \sqrt{x} + \sqrt{y} & = 5 \end{aligned}$

For integer solutions, $\Rightarrow \begin{cases} \sqrt{1} + \sqrt{16} = 5 \\ \sqrt{16} + \sqrt{1} = 5 \end{cases}$

Therefore, there are $\boxed{2}$ solutions.

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$\color{#3D99F6}{\text{Previous Solution}}$

For the equations to have integer solutions, both $x$ and $y$ must be perfect squares. For perfect square $x$ , we have:

$\begin{cases} x = 0 & \Rightarrow y = \left( \sqrt[3]{65} \right)^2 \ne \color{#D61F06}{\text{Not integer}} & \color{#D61F06}{\text{Rejected}} \\ x = 1 & \Rightarrow y = \left( \sqrt[3]{64} \right)^2 = 16 \quad \Rightarrow x\sqrt{y} + y\sqrt{x} = \color{#3D99F6} {20} & \color{#3D99F6}{\text{Accepted}} \\ x = 4 & \Rightarrow y = \left( \sqrt[3]{57} \right)^2 \ne \color{#D61F06}{\text{Not integer}} & \color{#D61F06}{\text{Rejected}} \\ x = 9 & \Rightarrow y = \left( \sqrt[3]{38} \right)^2 \ne \color{#D61F06}{\text{Not integer}} & \color{#D61F06}{\text{Rejected}} \\ x = 16 & \Rightarrow y = \left( \sqrt[3]{1} \right)^2 = 1 \quad \Rightarrow x\sqrt{y} + y\sqrt{x} = \color{#3D99F6} {20} & \color{#3D99F6}{\text{Accepted}} \end{cases}$

Therefore, there are $\boxed{2}$ integer solutions $(1,16)$ and $(16,1)$ .

Add the equations: $(x+y)(\sqrt{x}+\sqrt{y})=85=17\cdot 5$ .

So $\sqrt{x}+\sqrt{y}$ is an integer, so $x,y$ are perfect squares (simple proof here ).

$x,y\ge 1$ , so $x+y\ge \sqrt{x}+\sqrt{y}>1$ and $x+y=17, \sqrt{x}+\sqrt{y}=5$ .

$x,y\in\{1,4,9,16\}$ , which only gives $\{x,y\}=\{1,16\}$ . After checking, we see it is a solution.

$(x^{\frac 1 2 } + y^{\frac 1 2} )^3\\ = ( x \sqrt{x} + y \sqrt{y} ) + 3 ( x \sqrt{y} + y \sqrt{x} ) \\ =65+3*20=5^3 \\ \therefore~ x^{\frac 1 2 } + y^{\frac 1 2} =5=1+4=2+3.\\ ( x^{\frac 1 2 } , y^{\frac 1 2})~\text{(2,3) would not satisfy the given equations}\\ \text{where x and y are integers.}\\ \therefore~(1,16),(16,1) ~Two ~solutions. \\\text{CHECK:- Substityting (1,16) in the two given equations } ~~\\1\sqrt{16}+16\sqrt1 = 20,~~~~~~~~~~~~1\sqrt1+16\sqrt{16} =65.\\\text{Because of symmetry (16,1) also satisfy the given equations.}$

x=16,y=1 or vice versa

Step 1: find value of $\sqrt{x}+\sqrt{y}$ by adding triple of equation 1 to equation 2.

Step 2: factorize equation 1 to get $\sqrt{xy}=4$

Step 3:Let $\sqrt{x}=m,\sqrt{y}=n$ be roots of a quadratic whose sum and product is given as above. so m&n are roots of $x^2-5x+4=0$ Thus m,n=1,4 which means 2 solutions for m & n or 2 solutions for x & y.

This is also an easy solution

First note that $\sqrt{x}$ and $\sqrt{y}$ terms in the equation implies that $x \geq 0$ and $y \geq 0$

Consider $\begin{cases}x\sqrt{y} + y \sqrt{x} =20 \quad (1)\\ x\sqrt{x} + y\sqrt{y}=65 \quad (2) \end{cases}$

Squaring both equations and subtracting $(1)$ from $(2)$ we get $(x\sqrt{x} + y\sqrt{y})^2 -(x\sqrt{y} + y \sqrt{x})^2 =65^2 - 20^2$

$x^3+y^3-(x^2 y + y^2 x) =3825$

$(x+y)^3-4xy(x+y)=3825 \quad (\text{Writing} x^3+y^3=(x+y)^3-3xy(x+y))$

Put $x+y=S$ and $xy=P$

$S^3-4SP=3825 \quad (3)$

$S^3=3825+4P$

Clearly RHS is odd .

So $S$ must be odd and since $S^3 \geq 3825$ and just next perfect cube after $3825$ is $16^3=4096$ but $S$ cannot be even

If $S=17$ we get from eq $(3)$

$17^3-68P=3825\\ \implies 4913-68P=3825\\ \implies P=16\\ \implies x+y=17,xy=16 \implies (x,y)=(16,1),(1,16)$

Checking with the original equation these ordered pairs satisfy the given system.

Also notice that $S$ cannot be greater than $20$ as can be deduced from equation $(1)$ and as stated before $S$ cannot be even so we check the only case with $S=19$

If $S=19\\ S^3-4PS=3825\quad( \text{using eq.}(3))\\$ $19^3-76P=3825\\$ $6859-76P=3825\\$ $76P=3034\\$

$P=39.9210..$ which is not an integer.

So $S$ cannot be $19$

We conclude that the only solutions are

$(x,y)=(1,16),(16,1)$