TITLE: Find the number of values of $a$?
QUESTION [5 upvotes]: Consider a quadratic equation; 
$$ x^2 + 7x – 14(a^2 + 1) = 0,$$
… (where $a$ is an integer)
For how many different value of $a$, the equation will have at least one integer root?
I found out its discriminant, it comes out to be 
$$ (49 + 56(a^2+1))^{1/2}. $$ This should be the perfect square and also odd so that the at least one root be integer.
But I am unable to get the values.
How I can achieve this ?
Thanks in advance.

REPLY [3 votes]: If there is an integer solution, both (conceivably equal) solutions are integers, since their sum is $-7$.  Moreover, since $7$ divides the last two terms, it must divide any integer solution $x$. 
So let the solutions be $7a$ and $7b$. The product of the solutions is $49pq$. But it is also $-14(a^2+1)$. Now use Mark Bennet's calculation that shows that $a^2+1$ cannot be divisible by $7$.