**#Problem 7**

Show that there is no integer a such that a^2-3a-19 is divisible by 289

**.[R.M.O. 2009]**

**Solution Scheme and Approach**

a^2-3a-19

=a^2-3a-70+51

=(a+7)(a-10)+51 ...(i)

Now as (i) divides by 289 it should be divisible by 17 as well. So, it is clear at least one of (a+7) and (a-10) should be divisible by 17.

Let, a-10=17k =>a=10+17k therefore a+7=17(k+1)

Hence we can write a^2-3a-19=289k(k+1)+51

As 289k(k+1) is divisible by 289 so 51 has to be divisible by 289 as well..

Not possible, Hence Contradiction

**There is no integer a such that a^2-3a-19 is divisible by 289.Q.E.D.**

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