Question

Find the integer a such that $a^2-3a-19$ is divisible by $289$.

• A. $1$
• B. $2$
• C. $3$
• D. No integer possible

Answer 1

The correct option is D No integer possible

We write $a^2-3a-19=a^2-3a-70+51=(a-10)(a+7)+51$
Suppose 289 divides $a^2-3a-19$ for some integer a. Then 17 divides it and hence 17 divides (a-10)(a+7). Since 17 is a prime, it must divide (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 other also. Thus ${17}^2=289$ divides (a-10) and (a+7). It follows that 289 divides 51, which is impossible. Thus, there is no integer a for which 289 divides $a^2-3a-19$