卡塔兰猜想的一个特殊情况的证明

命题：\(x,y\)都是正整数，求证不存在\(x>2,y>3\)使得\(3^x-2^y=\pm 1\)

证：1° \(3^x-2^y = 1\)

\[ 3^x - 2^y = 1\]

\[\because x > 2,y > 3\\\]

\[\therefore 设s = x - 2,t = y - 3(s \ge 1,t \ge 1)\]

\[ 则3^2 \times 3^s - 2^3 \times 2^t = 1\\\]

\[ 9 \times 3^s-8 \times 2^t = 1\]

\[ \therefore 3^s = 1 + 8k,2^t = 1 + 9k (k \in Z)\]

\[ \because s \ge 1,t \ge 1\]

\[ \therefore k \ge 1\]

\[ \because 3^s = 1 + 8k （1）\]

\[ \therefore 3^s \equiv 1 \pmod 8\]

\[ 而当s = 2l(l \in Z)时，3^s \equiv 1 \pmod 8\]

\[\therefore 9^l = 8k + 1\]

\[ 又\because 2^t = 1 + 9k (2)\]

\[ 且当t = 6c (c \in Z)时，2^t \equiv 1 \pmod 9\]

\[ \therefore 64^c = 1 + 9k (3)\]

\[ 由(2),(3),有：\\ \begin{cases} k \equiv 1 \pmod 9 \\ k \equiv 7 \pmod {64}\\ \end{cases}\\ \]

\[所以k \equiv 199 \pmod{576}\]

\[ \therefore 9k + 1 \equiv 76 \pmod {576}\]

\[ 由(3) \\ 则64^c = 576r + 76(r \in Z)\]

\[ \because 64 | 576,64 \nmid 76\]

\[ \therefore 无解。 \]

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2° \(3^x - 2^y = -1\)

\[ 3^x - 2^y = -1\]

\[ 2^y - 3^x = 1\]

\[ 令s = y - 3,t = x - 2 \]

\[ 8 \times 2^s - 9 \times 3^t = 1\]

\[ 2^s = 8 + 9k,3^t = 7 + 8k (k\in Z)\]

\[ \because 3^t = 7 + 8k\]

\[ \therefore 3^t \equiv 7 \pmod 8\]

\[ 而3^t除8的余数集合为\{1,3\}\]

\[ \therefore 无解。\]

Q.E.D.

-----2020.1.18 20:50