指数方程解数

先看 $n=2$：

$$(x-y)(x+y)=2^{100}.$$

设 $x-y=2^a,\ x+y=2^b$，则

$$a+b=100,\ b>a,\ a\ge 1$$

（$a=0$ 时 $x,y$ 不为整数；$b>a$ 保证 $y>0$）。

所以 $a=1,2,\ldots,49$，共 49 组，且每组对应唯一正整数 $(x,y)$。

再证 $n>2$ 无解：

• 若 $n$ 为奇数，
  $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-1}).$$
  第二因子为大于 1 的奇数（在可行奇偶下），不可能使整体为纯 2 的幂。

• 若 $n=2m,\ m>1$，令 $u=x^m,v=y^m$，则
  $$u^2-v^2=2^{100}.$$
  即 $(u-v)(u+v)=2^{100}$，两因子均为 2 的幂。于是可写
  $$u=2^{a-1}(2^d+1),\ v=2^{a-1}(2^d-1).$$
  因为 $u,v$ 都是 $m$ 次幂且 $m>1$，可推出 $2^d-1$ 与 $2^d+1$ 都应为 $m$ 次幂，但它们相差 2，不可能同时是大于 1 的同次幂（Catalan 型结论），矛盾。

故只有 $n=2$ 有解，总解数为

$$\boxed{49}.$$

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英文解析

First consider $n=2$:

$$(x-y)(x+y)=2^{100}.$$

Let $x-y=2^a$ and $x+y=2^b$. Then

$$a+b=100,\ b>a,\ a\ge 1$$

(When $a=0$, $x$and $y$are not integers; $b>a$ensures $y>0$).

Thus $a=1,2,\ldots,49$, giving 49 groups, and each group corresponds to a unique positive integer pair $(x,y)$.

Next, prove there are no solutions for $n>2$:

• If $n$ is odd,
  $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+y^{n-1}).$$
  The second factor is an odd number greater than 1 (under feasible parity conditions), so it cannot make the whole expression a pure power of 2.

• If $n=2m$ with $m>1$, let $u=x^m$and $v=y^m$. Then
  $$u^2-v^2=2^{100}.$$
  This means $(u-v)(u+v)=2^{100}$, where both factors are powers of 2. Thus we can write
  $$u=2^{a-1}(2^d+1),\ v=2^{a-1}(2^d-1).$$
  Since $u$ and $v$ are both $m$-th powers and $m>1$, it follows that $2^d-1$and $2^d+1$must both be $m$-th powers. However, they differ by 2 and cannot simultaneously be $m$-th powers greater than 1 (a Catalan-type conclusion), which is a contradiction.

Therefore, only $n=2$ has solutions, and the total number of solutions is

$$\boxed{49}.$$