<s>[INST] Find the sum of all values of $a$ satisfying that there exist positive integers $a,b$ satisfying $(a-b) \sqrt{ab}=2016$ 
Find the sum of all values of $a$ satisfying that there exist positive integers $a,b$ satisfying $$(a-b) \sqrt{ab}=2016$$

my try 
let $a= x^2$  , $b= y^2$ $\to$ $(a-b) \sqrt{ab}$=$(x^2-y^2)(xy)=2016$
is this approach appropriate for this problem ? please elaborate your help  
 [/INST] A start of a solution. 
The prime factorization of $2016$ is $2^5\cdot3^2\cdot7$.
In order for $\sqrt{ab}$ to be a integer, both $a$ and $b$ must be multiples of squares and the same factors, i.e. $a=zx^2,b=zy^2, x>y$. Thus:
$$(a-b)\sqrt{ab}=z^2(x^2-y^2)xy = 2016$$
Use the fact that if neither of two numbers is divisible by 3, the difference of their squares must be. Thus, $z$ can not be a multiple of $3$, and since it cannot be a multiple of $7$, it must be one of $1,2,4$. 
So now we must solve:
$$(x^2-y^2)xy = 2016 = 2^5\cdot 3^2\cdot7$$
$$(x^2-y^2)xy = 504  = 2^3\cdot 3^2\cdot7$$
$$(x^2-y^2)xy = 126  = 2^1\cdot 3^2\cdot7$$
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