Vietnamese Mathematical Olympiad

Consider equation: $x^2 + 15y^2 = 4^n$ (*).
We have the following remarks:

Remark 1: If $(x, y)$ is a non-negative integer solution of (*) for $n = k$, $k \ge 1$, then $(2x, 2y)$ is a non-negative integer solution of (*) for $n = k + 1$.

Remark 2: For each $n \ge 2$, equation (*) always has 1 non-negative integer solution $(x, y)$, with $x, y$ odd. (Below such a solution will be called odd-solution).

Proof: We shall prove by induction in $n \ge 2$.
Since $1^2 + 15 \cdot 1^2 = 4^2$ the claim is valid for $n = 2$.
Assume that the claim is valid for $n = k$, $k \ge 2$, we will show its validity for $n = k + 1$.
Indeed, let $(x, y)$ be an odd-solution of (*) for $n = k$. Then, by means of the equality
$$4(x^2 + 15y^2) = \left(\frac{x + 15y}{2}\right)^2 + 15\left(\frac{x - y}{2}\right)^2 = \left(\frac{x - 15y}{2}\right)^2 + 15\left(\frac{x + y}{2}\right)^2$$
and $\frac{x + 15y}{2}, \frac{x - y}{2}, \frac{x - 15y}{2}, \frac{x + y}{2} \in \mathbb{Z}$
the pairs $\left(\frac{x + 15y}{2}, \frac{|x - y|}{2}\right)$ and $\left(\frac{|x - 15y|}{2}, \frac{x + y}{2}\right)$ are non-negative integer solution of (*) for $n = k + 1$.
Moreover, since $\frac{x - y}{2} + \frac{x + y}{2} = x$ is odd, one of the two integers $\frac{|x - y|}{2}$ and $\frac{x + y}{2}$ has to be odd. Consequently, one of the above two solution of (*) for $n = k + 1$ has to be odd. That is what to be proved.

Since $2^2 + 15 \cdot 0^2 = 4$, $(x = 2, y = 0)$ is a solution of (*) for $n = 1$. Hence, by means of Remarks 1 and 2, an easy induction show that for each $n \ge 1$, equation (*) always has at least $n$ non-negative integer solution $(x, y)$.