HKMO-2020 Math contest Question from math contest (already concluded in 2020) which I have tried but not yet found any elegant solution.
How many positive integer solutions does the following system of equations have?
$$\sqrt{2020}(\sqrt{a} + \sqrt{b} )= \sqrt{ (c+ 2020)(d+ 2020)}$$
$$\sqrt{2020}(\sqrt{b} + \sqrt{c} )= \sqrt{ (d+ 2020)(a+ 2020})$$
$$\sqrt{2020}(\sqrt{c} + \sqrt{d} )= \sqrt{ (a+ 2020)(b+ 2020)}$$
$$\sqrt{2020}(\sqrt{d} + \sqrt{a} )= \sqrt{ (b+ 2020)(c+ 2020)}$$
My attempt:
After few manipulations with an aim to remove the radicals, I end up with:
$$4040 \left(\sqrt{ab}+\sqrt{bc}+\sqrt{cd}+\sqrt{da}\right)=ab+bc+cd+da+4\cdot 2020^2$$
Before I go further, I feel that it is only going to become more unwieldy. Help is appreciated.
 A: Applying the Cauchy–Schwarz inequality to $(u_1,v_1,u_2,v_2) = (\sqrt{c},\sqrt{2020},\sqrt{2020},\sqrt{d})$ to the right hand side of the first equation
$$\sqrt{(c+2020)(2020+d)} \ge \sqrt{c}\sqrt{2020}+\sqrt{2020}\sqrt{d}= \sqrt{2020}(\sqrt{c}+\sqrt{d})$$
Then
$$\sqrt{a}+\sqrt{b}\ge \sqrt{c}+\sqrt{d} \tag{1}$$
Do the same method for the next 3 equation, we deduce
$$\sqrt{b}+\sqrt{c}\ge \sqrt{d}+\sqrt{a} \tag{2}$$
$$\sqrt{c}+\sqrt{d}\ge \sqrt{a}+\sqrt{b}\tag{3}$$
$$\sqrt{d}+\sqrt{a}\ge \sqrt{b}+\sqrt{c}\tag{4}$$
From $(1)(3)$ we deduce $$\sqrt{a}+\sqrt{b}= \sqrt{c}+\sqrt{d}$$
and the equality occurs when
$$\frac{c}{2020} = \frac{2020}{d}$$
$$\frac{a}{2020} = \frac{2020}{b}$$
Do the same thing, from $(2)(4)$, we must have
$$\frac{b}{2020} = \frac{2020}{c}$$
$$\frac{d}{2020} = \frac{2020}{a}$$
Hence, we have
$$\frac{a}{2020}=\frac{2020}{b}=\frac{c}{2020}=\frac{2020}{d} $$
or
$$(a,b,c,d)=\left(t,\frac{2020^2}{t},t,\frac{2020^2}{t}  \right) \tag{5}$$
From $(5)$, we have $t| 2020^2 = (2^4 5^2 101^2)$ So, there are $(4+1)(2+1)(2+1)=45$ different values of $t$.
Conclusion: there are $45$ solutions
$$(a,b,c,d)=\left(t,\frac{2020^2}{t},t,\frac{2020^2}{t}  \right) \qquad \forall t| 2020^2$$
