KVPY SA 2016

With $a,b,c \gt 0$ in arithmetic progression, let $b = a + k$ and $c = a + 2k$ for some real $k$ . Then by Vieta's we know that $P + Q = -\dfrac{b}{a} = -1 - \dfrac{k}{a}$ and that $PQ = \dfrac{c}{a} = 1 + 2\dfrac{k}{a}$ , and thus $P + Q + PQ = \dfrac{k}{a}$ .

Now as $P,Q$ must be integers we will require that $\dfrac{k}{a}$ be an integer, and so $k = na$ for some integer $n$ . It is the possible values of $n$ that we are then wanting to find. The given quadratic can then be written as

$ax^{2} + (a + na)x + (a + 2na) = a(x^{2} + a(n + 1)x + a(2n + 1)) = 0 \Longrightarrow x^{2} + (n + 1)x + (2n + 1) = 0$

$\Longrightarrow x = \dfrac{-(n + 1) \pm \sqrt{(n + 1)^{2} - 4(2n + 1)}}{2} = \dfrac{-(n + 1) \pm \sqrt{n^{2} - 6n - 3}}{2}$ ,

which will only have integer solutions if $n^{2} - 6n - 3 = m^{2}$ for some (positive) integer $m$ . Completing the square, we then have that

$(n - 3)^{2} - 12 = m^{2} \Longrightarrow (n - 3)^{2} - m^{2} = 12 \Longrightarrow ((n - 3) - m)((n - 3) + m) = 12$ .

Now when $12$ can be factored as $ab$ , with $a \lt b$ , then we can assign $a = (n - 3) - m$ and $b = (n - 3) + m$ , for which $a + b = 2(n - 3)$ , an even integer. So we need only consider the factorings of $12$ where either $a,b$ are both even or are both odd. These are $(a,b) = (2,6)$ and $(a,b) = (-6,-2)$ , which give us values for $n$ of $7$ and $-1$ . With $n = -1$ we would end up with $b = 0$ , violating the condition that $b$ be positive, so we are left with $P + Q + PQ = n = \boxed{7}$ .

For this value of $n$ we have that $m = 2$ , and thus $x = \dfrac{-8 \pm 2}{2} \Longrightarrow P = -3, Q = -5$ . So for example, with $a = 1$ we would have $b = 8$ and $c = 15$ , giving us the quadratic $x^{2} + 8x + 15 = (x + 3)(x + 5)$ , which has integer roots $P = -3, Q = -5$ such that $P + Q + PQ = -3 + (-5) + (-3)(-5) = 7$ as anticipated.