Hong Kong (China) Mathematical Olympiad

$n = 1$.
When $n = 1$, $(2^1 - 1)(5^1 - 1) = 4$ is a perfect square. Suppose $n > 1$.

* If $n \equiv 3 \pmod{4}$, we have $(2^n - 1)(5^n - 1) \equiv (2)(-1) \equiv 3 \pmod{5}$, which is not true for a square.

* If $n \equiv 1 \pmod{4}$, we rewrite
$$(2^n - 1)(5^n - 1) = (2^n - 1) \cdot 4(5^{n-1} + 5^{n-2} + \dots + 1).$$
If this is a square, so is $(2^n - 1)(5^{n-1} + 5^{n-2} + \dots + 1)$. However,
$$(2^n - 1)(5^{n-1} + 5^{n-2} + \dots + 1) \equiv (-1)(1 + 1 + \dots + 1) = -n \equiv 3 \pmod{4},$$
which is not true for a square.

* If $n = 2m$ is even, suppose $(2^n - 1)(5^n - 1)$ is a perfect square. Let $2^n - 1 = ds^2$ and $5^n - 1 = dt^2$ where $d$ is squarefree. We first show that $d > 1$. Indeed, if $d = 1$, then $5^{2m} - t^2 = 1$. This gives $(5^m - t)(5^m + t) = 1$, which is impossible as $t > 0$. Now, it is known that the solutions to Pell's equation $x^2 - dy^2 = 1$ are generated by $(x_1 + y_1\sqrt{d})^n = x_n + y_n\sqrt{d}$, where $(x_1, y_1)$ is the fundamental solution.
Claim. Let $(x_k, y_k)$ be a solution to $x^2 - dy^2 = 1$.
(i) If $k \equiv 0 \pmod{2}$, then $x_k \equiv 1 \pmod{2}$ and $y_k \equiv 0 \pmod{2}$.
(ii) If $k \equiv 1 \pmod{2}$, then $x_k \equiv x_1 \pmod{2}$ and $y_k \equiv y_1 \pmod{2}$.

Proof.
(i) Let $k = 2l$. Then we have
$$\begin{aligned} x_k + y_k \sqrt{d} &= (x_1 + y_1 \sqrt{d})^{2l} \\ &= (x_l + y_l \sqrt{d})^2 \\ &= x_l^2 + dy_l^2 + 2x_l y_l \sqrt{d} \\ &= 2x_l^2 - 1 + 2x_l y_l \sqrt{d}. \end{aligned}$$
So $x_k = 2x_l^2 - 1 \equiv 1 \pmod{2}$ and $y_k = 2x_l y_l \equiv 0 \pmod{2}$.

$$x_{2l+1} = x_{2l}x_1 + y_{2l}y_1 d \equiv x_1 \pmod{2},$$
$$y_{2l+1} = x_{2l}y_1 + x_1 y_{2l} \equiv y_1 \pmod{2}.$$
$\Box$
Now, since $2^{2m} - 1 = ds^2$, $(x, y) = (2^m, s)$ is a solution to $x^2 - dy^2 = 1$. By the claim, we must have $x_1 \equiv 0 \pmod{2}$. Similarly, since $5^{2m} - 1 = dt^2$, $(x_k, y_k) = (5^m, t)$ is another solution. By the claim, as $5^m \not\equiv x_1 \pmod{2}$, we must have $k \equiv 0 \pmod{2}$. Let $k = 2l$. Then we have $5^m = x_{2l} = 2x_l^2 - 1$. This implies $2x_l^2 - 1 \equiv 0 \pmod{5}$, and hence $x_l^2 \equiv 3 \pmod{5}$. This is impossible.
Therefore, the only solution is $n = 1$.