是
by __Tonycyt__ @ 2024-04-20 12:06:12
$b=1$ 的话不是有无数个 $y$ 么（
by vflower @ 2024-04-20 12:23:10
@[vflower](/user/1000066) 忘记说了，$a,b>1$
by Albert_Wei @ 2024-04-20 12:28:16
@[__Tonycyt__](/user/667037) 证明？
by Albert_Wei @ 2024-04-20 12:32:18
@[Albert_Wei](/user/676634) 我不太记得，我以前数学做过这道题，现在只记得结论了qwq
by __Tonycyt__ @ 2024-04-20 12:47:17
@[__Tonycyt__](/user/667037) 好吧/kk
by Albert_Wei @ 2024-04-20 12:54:58
我想到一种貌似合理的解法：
设存在两组解 $(x_0,y_0),(x_1,y_1),x_0<x_1$。显然 $y_0<y_1$。
$$a^{x_0}-b^{y_0}=c
\\
a^{x_0}=c+b^{y_0}
$$
因此
$$
a^{x_1}=a^{x_0} a^{x_1-x_0}
\\
=(c+b^{y_0})a^{x_1-x_0}
$$
$$
a^{x_1}-b^{y_1}=c\\
(c+b^{y_0})a^{x_1-x_0}-b^{y_1}=c\\
a^{x_1-x_0}=\frac{c-b^{y_1}}{c+b^{y_0} }<1
$$
因此 $x_1=x_0$。
by txppdd @ 2024-04-20 14:38:29
@[txppdd](/user/542128) 那么即使 $x_1=x_0$，$a^{x_1-x_0}<1$ 也不成立啊
by __Tonycyt__ @ 2024-04-20 15:16:14
你的解法是错的：$a^{x_1-x_0}=\dfrac{c+b^{y_1}}{c+b^{y_0}}>1$
by __Tonycyt__ @ 2024-04-20 15:18:03
@[Albert_Wei](/user/676634) 怎么这么像比尔猜想
by Link_Cut_Y @ 2024-04-20 16:43:50