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Chapter 4: Problem 8

Let \(a, b, c \in \mathbf{Z}^{+}\)with \(\operatorname{gcd}(a, b)=1\). If \(a \mid c\) and \(b \mid c\), prove that \(a b \mid c\). Does the result hold if \(\operatorname{gcd}(a, b) \neq 1\) ?

Short Answer

Expert verified
Yes, the product \(ab\) does divide \(c\) when \(a\) and \(b\) are coprime and both divide \(c\), but the result does not hold when gcd(a, b) is not equal to 1.

Step by step solution

01

Understand the conditions

We are given that \(a, b, c \in \mathbf{Z}^{+}\), which means a, b, and c are positive integers. Also, \(\operatorname{gcd}(a, b)=1\), meaning that a and b are coprime or relatively prime, i.e., the largest integer that divides both of them is 1. Further, we have that a divides c and b divides c. It means there exist integers \(x\) and \(y\) such that \(ax = c\) and \(by = c\). The goal is to prove that the product of a and b divides c. In other words, we need to prove that there exists an integer \(z\) such that \(abz = c\).
02

Apply Bezout's lemma

We can apply Bezout's lemma, which states that if \(a\) and \(b\) are coprime, there exist integers \(p\) and \(q\) such that \(ap + bq = 1\). Multiplying through by \(c\), we obtain \(apc + bqc = c\). From the divisibility conditions, we know that there exist integers \(x\), \(y\) such that \(ax = c\) and \(by = c\). Substituting these into the equation, we get \(ap(ax) + bq(by) = c\). This simplifies to \(ab(px + qy) = c\). Hence we have shown that \(ab\) divides \(c\).
03

Consider the case gcd(a, b) not equal to 1

If gcd(a, b) is not equal to 1, the assertion does not necessarily hold true. For example, if \(a = 2\), \(b = 4\) and \(c = 8\), then clearly \(a\) divides \(c\) and \(b\) divides \(c\), but their product (8) does not divide \(c\). So the statement only holds when \(a\) and \(b\) are coprime (gcd(a, b) = 1).

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Most popular questions from this chapter

As in Example \(4.20\) let \(L_{0}, L_{1}, L_{2}, \ldots\) denote the Lucas numbers, where (1) \(L_{0}=2, L_{1}=1\); and (2) \(L_{n+2}=L_{n+1}+\) \(L_{n}\), for \(n \geq 0\). When \(n \geq 1\), prove that $$ L_{1}^{2}+L_{2}^{2}+L_{3}^{2}+\cdots+L_{n}^{2}=L_{n} L_{n+1}-2 $$

Let \(n \in \mathbf{Z}^{+}\)with \(u\) the units digit of \(n\). Prove that \(7 \mid n\) if and only if \(7 \mid\left(\frac{n-u}{10}-2 u\right)\).

Let \(n \in \mathbf{Z}^{+}\)with \(n=r_{k} \cdot 10^{k}+\cdots+r_{2} \cdot 10^{2}+r_{1} \cdot 10+r_{0}\) (the base- 10 representation of \(n\) ). Prove that a) \(2 \mid n\) if and only if \(2 \mid r_{0}\) b) \(4 \mid n\) if and only if \(4 \mid\left(r_{1} \cdot 10+r_{0}\right)\) c) \(8 \mid n\) if and only if \(8 \mid\left(r_{2} \cdot 10^{2}+r_{1} \cdot 10+r_{0}\right)\) State a general theorem suggested by these results.

If \(a, b \in \mathbf{Z}^{+}\), and both are odd, prove that \(2 \mid\left(a^{2}+b^{2}\right)\) but \(4 \chi\left(a^{2}+b^{2}\right)\)

For \(a, b, n \in \mathbf{Z}^{+}\), prove that \(\operatorname{gcd}(n a, n b)=n \operatorname{gcd}(a, b)\).
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