# Elementary Number Theory Problems 3.3 Solution (David M. Burton's 7th Edition) - Q21

My Solution for "(a) For any integer $k > 0$, establish that the arithmetic progression $$a + b, a + 2b, a + 3b, ...$$ where $gcd(a, b) = 1$, contains $k$ consecutive terms that are composite. [Hint: Put $n =(a+ b)(a + 2b) \cdots (a+ kb)$ and consider the $k$ terms $a+ (n + 1)b, a+ (n + 2)b, ..." ## Table of Contents ## Background All theorems, corollaries, and definitions listed in the book's order: I will only use theorems or facts that are proved before this question. So you will not see that I quote theorems or facts from the later chapters. ## Question (a) For any integer$k > 0$, establish that the arithmetic progression $$a + b, a + 2b, a + 3b, ...$$ where$gcd(a, b) = 1$, contains$k$consecutive terms that are composite. [Hint: Put$n =(a+ b)(a + 2b) \cdots (a+ kb)$and consider the$k$terms$a+ (n + 1)b, a+ (n + 2)b, ... , a+ (n + k)b$.] (b) Find five consecutive composite terms in the arithmetic progression $$6, 11, 16,21,26,31,36, ...$$ ## Solution ### (a) Let$i$be an integer where$i \in {1, 2, 3, .... k}$. Put$n =(a + b)(a + 2b) \cdots (a+ kb)$and let us consider the$k$terms$a + (n + 1)b, a + (n + 2)b, ... , a + (n + k)b$. For any$a + (n + i)b$term: $$\begin{equation} \begin{split} a + (n + i)b & = a + [(a + b)(a + 2b) \cdots (a+ kb) + i]b \\ & = a + ib + (a + b)(a + 2b) \cdots (a+ kb)b \\ & = (a + ib) + (a + b) \cdots (a + (i - 1)b)(a + ib)(a + (i + 1)b) \cdots (a+ kb)b \\ & = (a + ib)[1 + (a + b) \cdots (a + (i - 1)b)(a + (i + 1)b) \cdots (a+ kb)b] \end{split} \nonumber \end{equation}$$ Therefore, every$(a + (n + i)b)$term is divisible by$(a + ib)$, and they are$k$consecutive terms. So, the arithmetic progression contains$k$consecutive terms that are composite. ### (b) The arithmetic progression is the form$a + mb$where$a = 1$and$b = 5$with$m \in Z^{+}$. From part (a), we have$n = (6)(11)(16)(21)(26) = 576576$. The terms are:$1 + (576576 + 1)(5) = 2882886 = (1 + 5 \times 1)(480481)1 + (576576 + 2)(5) = 2882891 = (1 + 5 \times 2)(262081)1 + (576576 + 3)(5) = 2882896 = (1 + 5 \times 3)(180181)1 + (576576 + 4)(5) = 2882901 = (1 + 5 \times 4)(137281)1 + (576576 + 5)(5) = 2882906 = (1 + 5 \times 5)(110881)\$

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