Number's application

#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.