3. let $n$ be a natural number. Determine the number of pairs $(a, b)$ of natural numbers for which the following equation is satisfied:

$$
(4 a-b)(4 b-a)=2010^{n}
$$

## Solution
Set $x=4 a-b$ and $y=4 b-a$. Because $x y>0$ and $x+y=3(a+b)>0$, $x$ and $y$ are natural numbers. Furthermore

$$
a=\frac{4 x+y}{15}, \quad b=\frac{4 y+x}{15}
$$

so $4 x+y$ and $4 y+x$ must both be divisible by 15. However, because of $15 \mid x y$, at least one of the two numbers $x, y$ is divisible by 3 and at least one of the two is divisible by 5. Thus, both must be divisible by 15 and in this case $a, b$ are actually whole. These considerations provide a bijection between the set of all pairs $(a, b)$ that fulfill the above equation and the set $S$ of all pairs $(x, y)$ of natural numbers with the properties

- $15 \mid x$ and $15 \mid y$.
- $x y=2010^{n}$.

Because $2010=2 \cdot 3 \cdot 5 \cdot 67$, $x$ is of the form $x=2^{a} \cdot 3^{b} \cdot 5^{c} \cdot 67^{d}$ with $0 \leq a, d \leq n$ and $1 \leq b, c \leq n-1$ and $y=2010^{n} / x$. So in total there are

$$
(n+1)^{2}(n-1)^{2}=\left(n^{2}-1\right)^{2}
$$

Solutions to this equation.