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1. Calculation: $ (5767, 4453) $, $ (3141, 1592) $, $ (136, 221, 391) $.
Answer: $$ (5767, 4453) = (4453, 1314) = (1314, 511) = (511, 292) = (292, 219) = (219, 73) = 73.$$ $$ (3141, 1592) = (1 592, 1549) = (1549, 43) = (43, 1) = 1.$$ $$ (136, 221, 391) = (136, (221, 391)) = (136, (221, 170)) = (136, 170, 51) 36, 17) = 17.$$
2. Set $n $ is an integer that proves: $$ (12n + 5, 9n + 4) = 1.$$ solution:
By Bezout Identity, $$4 (9n+4)-3 (12n+5) = 1\rightarrow (12n + 5, 9n + 4) = 1.$$
3. Set $m, n\in\mathbf{n^*}$, and $m $ is odd, proving: $$\left (2^m-1, 2^n + 1\right) = 1.$$ Solution:
Set $d = \left (2^m-1, 2^n + 1\right) \rightarrow 2^m-1 = da$, $2^n + 1 = db$, $ (A, B) = 1$. $$\rightarrow 2^{mn} = (da + 1) ^n = (db-1) ^m$$ $$\rightarrow da + 1 = db-1\rightarrow D (b-a) = 2 \rightarrow d\ |\ 2 \ RightArrow d = 1, 2.$$ easy to know $d $ is odd, so $d = 1$.
4. Set $ (A, B) = 1$, proof: $$\left (a^2 + b^2, ab\right) = 1.$$ solution: $$ (A, b) = 1\rightarrow (a^2, b) = 1 \rig Htarrow (a^2 + b^2, b) = 1.$$ similarly available $ (a^2 + b^2, a) = 1$. So $ (a^2 + b^2, AB) = 1$.
5. Proof: If $a, b\in\mathbf{n^*}$, then sequence: $a $, $2a$, $\cdots$, $ba $ The number of items that can be divisible by $b $ equals $ (A, b) $.
Answer:
Make $ (A, B) = d$, $a = da_1$, $b = db_1$, $ (a_1, b_1) = 1$. The original number is listed as $ $da _1, da_2, \cdots, db_1da_1,$$ at the same time divided by $b $ $$\frac{a_1}{b_1}, \frac{2a_1}{b_1}, \cdots, \frac{(db_1) a_1}{b_1},$$ its The items in the integer are $\dfrac{(ib_1) a_1}{b_1}$, where $i = 1, 2, \cdots, and d$ total $d $.
2017 Winter Monkey Tutoring Elementary Number theory-2: "With more Division" work problem answer