Solutions pairs for the equation $(4a - b) (4b - a) = 1770n$ 
How many solution pairs $(a, b)$ of positive integers exist for the equation
  $(4a - b) (4b - a) = 1770n$,
  when $n$ is a positive integer?

This one seemed to be pretty tough, it's from Sweden's national contest. How should one approach this?
 A: The main crux is that $4^2 - 1 \mid 1770$.
You can modify this solution to solving $ (ka-b)(kb-a) = (k^2-1) n$ over the integers, and showing that solutions exist iff $ (k^2 - 1) \mid n$.   

Hint: Show that $ 15 \mid 4a-b, 15 \mid 4b-a$.
Note: This might seem artificial, but it's a very natural result when you start playing with the problem. 

 Let $ c = 4a-b, d = 4b-a$.
 For $ a = \frac{ 4c+d}{15} , b = \frac{ 4c+d}{15} $ to be integers, we require $ c \equiv - 4d \pmod{15}$, and thus $ 0 \equiv 1770 = cd \equiv - 4d^2 \pmod{15}.$
 Since 15 is square-free, thus $15 \mid d$. Likewise, $15 \mid c$.     


Corollary: Thus, $15^2 \mid (4a-b)(4b-a)$. So if $ 15 \not \mid n$, then there are no solutions.   

Conversely, if $ n = 15 k$, 
Hint: Relax the condition to allow for negative $a, b$ for now. Then, the complete set of integer solutions can be described as:   

  Let $e,f$ be any pair of integers whose product is $118k$.
 Then $a = 4e+f, b = 4f+e$.   

Clearly, if $e, f$ are both negative, then $a, b$ are also both negative, and we should reject.
If $e, f$ are both positive, then $a, b$ are also both positive.
Hence, the complete set of positive integer solutions is

  Let $e,f$ be any pair of positive integers whose product is $118k$. Then $a = 4e+f, b = 4f+e$.
 The number of solutions is equal to the number of divisors of $\frac{ 118n } { 15} $.   

E.g. When $n = 15$, we get the solutions $ (a,b) = (67, 238), (238, 67), (122, 473), (473, 122)$ which arise from $ef = 118$,  $(e,f) = (2, 59), (59, 2), (1, 118), (118, 1)$.
