Trust your identities 2

Note that $(a+b+c)(a+b-c) = (a+b)^2 - c^2 = a^2 + 2ab + b^2 - c^2$ . Let $a=x$ , $b=1$ , and $c=y$ , Then we have:

$\begin{aligned} (x+1+y)(x+1-y) & = \blue{x^2 + 2x} + 1 - \blue{y^2} & \small \blue{\text{Since }x^2 + 2x - y^2 = 20} \\ & = \blue {20} + 1 = 21 \end{aligned}$

Since $x$ and $y$ are integers and $x > y$ . this means that $x+1 +y > x+1 - y$ . Since $(x+1+y)(x+1-y) = 21 = 7 \times 3$ , we can assume that: $x+1+y = 7$ and $x+1 - y = 3$ . From $x+1+y = 7 \implies x+y = \boxed 6$ .