Let a and b be coprime integers, and



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Let a and b be coprime integers, and let m be an integer such that a | m and b | m. Prove that ab | m

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Expert2023-04-27Added 556 answers

To prove that abm, we need to show that m is a multiple of ab, which means that m=abn for some integer n.
Since am, we can write m=ak for some integer k. Similarly, since bm, we can write m=bl for some integer l.
Since a and b are coprime, we have gcd(a,b)=1. Therefore, there exist integers x and y such that ax+by=1 (this is the Bezout's identity).
Multiplying both sides by m, we get (ax+by)m=m. Since m=ak=bl, we can substitute these values to get (ax+by)ak=(ax+by)bl.
Expanding both sides, we get axak+byak=axbl+bybl, which simplifies to a2xk+abyk=abxl+b2yl.
Since am and bm, we know that m is a common multiple of a and b. Therefore, ab is a common multiple of a and b.
Using the definition of divisibility, we can say that aba2 and abb2. Hence, ab divides the left-hand side of the above equation.
Since aba2xk, ababyk, and ababxl, we can conclude that abb2yl.
Since a and b are coprime, b does not divide a. Therefore, b must divide yl, which means that there exists an integer n such that yl=bn.
Substituting this value, we get abb2bn, which simplifies to abn=m.
Hence, we have shown that if a and b are coprime integers, and if m is an integer such that am and bm, then abm.

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