prmo 2013 question 3

3. It is given that the equation 

$x^2 + ax + 20 = 0$ has integer roots. What is the sum of all possible values of $a$?
[PRE-RMO – 2013]

Solution 1:

x = (a ± √(a² - 80)) / 2

a² - 80 should be a perfect square.

For eg. a = 9 ⇒ √1

a = -9 also works

And x should be integer.

So numerator, i.e. a ± √(a² - 80) should be even.

If a is odd, (a² - 80) is odd and hence

√(a² - 80) is odd

⇒ numerator = -a ± odd = -odd ± odd = even

Similarly if a is even, (a² - 80) is even and hence

√(a² - 80) is even

⇒ numerator = -a ± even = -even ± even = even

So numerator is always even if (a² - 80) is a perfect square.
And for each valid value of (a² - 80), there are 2 valid values of 'a'(plus and minus).
So no matter how many valid values of 'a' are there, they will all add up to 0.
Answer: 0

Solution 2:

A convenient way to proceed is to let the integer roots be 

$r_1$ and $r_2$. Then by Vieta’s formulas,

$$r_1 + r_2 = -a \quad\text{and}\quad r_1\,r_2 = 20.$$

Since both $r_1$ and $r_2$ must be integers whose product is 20, the only possibilities (up to order) are

$$(1, 20),\,(2, 10),\,(4, 5),\,(-1, -20),\,(-2, -10),\text{ or }(-4, -5).$$

Their respective sums (and corresponding values of $-a$) are

• $1+20=21\implies a=-21$

• $2+10=12\implies a=-12$

• $4+5=9\implies a=-9$

• $(-1)+(-20)=-21\implies a=21$

• $(-2)+(-10)=-12\implies a=12$

• $(-4)+(-5)=-9\implies a=9.$

Hence the possible values of $a$ are $\{-21, -12, -9, 9, 12, 21\}$. Their sum is

$$(-21) + (-12) + (-9) + 9 + 12 + 21 \;=\; 0.$$

Therefore, the sum of all possible values of $a$ is $\boxed{{\displaystyle 0}}\mathrm{.}$