Show that there is no integer a such that a2 – 3a – 19 is divisibl

1.	Show that there is no integer a such that a2 – 3a – 19 is divisible by 289.
Answer» Explanation: We write a² - 3a - 19 = a² - 3a - 70 + 51 = (a - 10)(a + 7) + 51 Suppose 289 divides a² - 3a - 19 for some integer a. Then 17 divides it and hence 17 divides (a - 10)(a + 7). Since 17 is a prime, it must divides (a - 10) or (a + 7). But (a + 7) - (a - 10) = 17. Hence whenever 17 divides one of (a - 10) and (a + 7), it must divide the older also. Thus 17² = 289 divides (a - 10)(a + 7). It follows that 289 divides 51, which is impossible. Thus, there is no integer a for which 289 divides a² - 3a - 19.

Show that there is no integer a such that a2 – 3a – 19 is divisible by 289.

Answer»

Explanation:

We write

a² - 3a - 19 = a² - 3a - 70 + 51 = (a - 10)(a + 7) + 51

Suppose 289 divides a² - 3a - 19 for some integer a. Then 17 divides it and hence 17 divides (a - 10)(a + 7). Since 17 is a prime, it must divides (a - 10) or (a + 7). But (a + 7) - (a - 10) = 17. Hence whenever 17 divides one of (a - 10) and (a + 7), it must divide the older also. Thus 17² = 289 divides (a - 10)(a + 7). It follows that 289 divides 51, which is impossible. Thus, there is no integer a for which 289 divides a² - 3a - 19.

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