Find the number of values of $a$? 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.
 A: So, the discriminant $(D^2)$ is $49+56(a^2+1)=7(8a^2+15)$
As $7\mid D^2\implies 7\mid D$ if $D$ is an integer,
in that case, $49\mid D^2$ i.e., $49\mid \{49+56(a^2+1)\}$
$\implies 7\mid (8a^2+15)\implies 7\mid (a^2+1)$
$-1$ needs to be  a quadratic residue of $7$.
Using Euler's Criterion,
$-1$ is a quadratic residue of primes $\equiv 1\pmod 4$, but $7\equiv -1\pmod 4$,
Without using quadratic residue, all the integers can be written as $7b,7b\pm1,7b\pm2,7b\pm 3$
So,  $(7b)^2\equiv 0\pmod 7,(7b\pm1)^2\equiv 1\equiv -6,$
$(7b\pm 2)^2\equiv 4\equiv -3, (7b\pm3)^2\equiv 2\equiv -5$,
 so there is no solution to $a^2\equiv -1\pmod 7$
So, there is no rational solution of the given equation for the integral values of $a$.
A: By Eisenstein's criterion we need $a^2+1$ to be divisible by 7 for the equation to have a rational root. The squares mod 7 are 0,1,2,4, and do not include -1, so this cannot happen.
A: 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 $7$.
