Find the number of ordered pairs!

Number Theory Level 4

$\large x^2 - px + q = 0 \qquad \qquad x^2-qx +p = 0$

Let $p$ and $q$ be positive integers such that the two quadratic equations above have unequal integer roots (in $x$ ).

Find the number of ordered solutions $(p,q)$ satisfying the above conditions.

1 solution

Ishan Singh
May 16, 2016

Let roots of $x^2-px+q=0$ are $a$ and $b$ . Note that both roots are positive integers.

Since the roots of the two equations are unequal, W.L.O.G. $p > q$

Now,

$a+b = p$

$ab=q$

$\implies a+b>ab \ (\because p>q)$

$\implies (a-1)(1-b) > -1$

Since $a,b \in \mathbb{Z^+}$ , we have,

$(a-1)(1-b) \leq 0$

$\implies -1<(a-1)(1-b) \leq 0$

$\implies a=1 \ \text{or} \ b=1$

$\implies q=p-1$

From the second equation, for integral roots, discriminant must be a perfect square. Thus,

$q^2 - 4p = k^2 \ ; \ k \in \mathbb{Z}$

$\implies p^2-6p+1=k^2$

$\implies (p-3+k)(p-3-k) = 8$

$\implies p-3+k=4 \ ; \ p - 3 - k = 2$

$\implies k=1, \ p=6, \ q=5$

Thus, solutions are $(6,5)$ and $(5,6)$

$\therefore$ Number of solutions $= \boxed{2}$