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