CAT 2019 Slot 2 — QA Question 22

Factor as difference of squares. Move $m^{2}$ across and factor.

$$\begin{aligned} &n^{2}-m^{2}=105\\ &\Rightarrow\ (n-m)(n+m)=105 \end{aligned}$$

List factor pairs of 105. Since $105=3\times5\times7$ is odd, every factor is odd, so each unordered pair $(d,\tfrac{105}{d})$ with $d<\tfrac{105}{d}$ gives integer $n=\tfrac{(n+m)+(n-m)}{2}$ and $m=\tfrac{(n+m)-(n-m)}{2}$.

$$\begin{aligned} &105=1\times105=3\times35=5\times21=7\times15 \end{aligned}$$

Solve each pair. Set $n-m$ to the smaller factor, $n+m$ to the larger; all give positive integers.

$$\begin{aligned} &(1,105)\Rightarrow (m,n)=(52,53)\\ &(3,35)\Rightarrow (m,n)=(16,19)\\ &(5,21)\Rightarrow (m,n)=(8,13)\\ &(7,15)\Rightarrow (m,n)=(4,11) \end{aligned}$$