UKMT Senior Challenge-III

Let the two roots of the equation $x^2+ax+2013=0$ are $\alpha$ , $\beta$

Then the equation becomes

$x^2-(\alpha+\beta)x+\alpha\beta=0$

$x^2-(\alpha+\beta)x+2013=0$ [Replacing $\alpha\beta$ by $2013$ ]

Now as $2013=3*11*61$

We get the solutions of $\alpha+\beta=a$ ${(1+2013=2014),(3+11*61=674),(3*11+61=94),(3*61+11=194)}$

Thus $4$ solutions.All can be negative so the total number of solutions will be $4*2=\boxed{8}$

For x to be an integer we the discriminant to be a square: $\ b^2 -4ac = a^2 -4(2013) = b^2 \Rightarrow\ a^2 -b^2 = (a-b)(a+b) = 2^2 \times\ 3 \times\ 11 \times\ 61$ Now the sum of the 2 factors, a-x and a+x must be even (namely 2a) and at least one of them is even. However, if say, a-x is odd and a+x is even then this implies (by parity) that 2a is odd. This contradiction proves that we pick a single factor of 2 for a-x and a+x. We have 2 choices for each of the others leaving: $\ 1 \times\ 2 \times\ 2 \times\ 2 = \boxed{8} \ ~ \ choices$

If we write x(x+a)=-2013 and we know that 'a' and 'x' must be integers then we can simply just count the amount of integer factors that 2013 has. This is 8 (1, 61, 183, 671, 3, 11, 33, 2013). We find these by permutations of the prime factors of 2013.