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1. 计算: $(5767, 4453)$, $(3141, 1592)$, $(136, 221, 391)$.
解答: $$(5767, 4453) = (4453, 1314) = (1314, 511) = (511, 292) = (292, 219) = (219, 73) = 73.$$ $$(3141, 1592) = (1592, 1549) = (1549, 43) = (43, 1) = 1.$$ $$(136, 221, 391) = (136, (221, 391)) = (136, (221, 170)) = (136, (170, 51)) = (136, 17) = 17.$$
2. 设 $n$ 是整数, 证明: $$(12n + 5, 9n + 4) = 1.$$ 解答:
由Bezout恒等式, $$4(9n+4) - 3(12n+5) = 1Rightarrow (12n + 5, 9n + 4) = 1.$$
3. 设 $m, ninmathbf{N^*}$, 且 $m$ 是奇数, 证明: $$left(2^m - 1, 2^n + 1 ight) = 1.$$ 解答:
设 $d = left(2^m - 1, 2^n + 1 ight)Rightarrow 2^m - 1 = da$, $2^n + 1 = db$, $(a, b) = 1$. $$Rightarrow 2^{mn} = (da + 1)^n = (db - 1)^m$$ $$Rightarrow dA + 1 = dB - 1Rightarrow d(B-A) = 2 Rightarrow d | 2 Rightarrow d = 1, 2.$$ 易知 $d$ 是奇数, 因此 $d = 1$.
4. 设 $(a, b) = 1$, 证明: $$left(a^2 + b^2, ab ight) = 1.$$ 解答: $$(a, b) = 1Rightarrow (a^2, b) = 1 Rightarrow (a^2 + b^2, b) = 1.$$ 同理可得 $(a^2 + b^2, a) = 1$. 因此 $(a^2 + b^2, ab) = 1$.
5. 证明: 若 $a, binmathbf{N^*}$, 那么数列: $a$, $2a$, $cdots$, $ba$ 中能被 $b$ 整除的项的个数等于 $(a, b)$.
解答:
令 $(a, b) = d$, $a = da_1$, $b = db_1$, $(a_1, b_1) = 1$. 则原数列为 $$da_1, da_2, cdots, db_1da_1,$$ 同时除以 $b$ 得 $$frac{a_1}{b_1}, frac{2a_1}{b_1}, cdots, frac{(db_1)a_1}{b_1},$$ 其中为整数的项为 $dfrac{(ib_1)a_1}{b_1}$, 其中 $i = 1, 2, cdots, d$ 总计 $d$ 个.