Using \((1) – (2)\), we get \[\begin{aligned} x^{2}-y^{2} &=(4 \sqrt{5})^{2}-(4 \sqrt{2})^{2} \\\\ &=80-32 \\\\ &=48 \\\\ (x-y)(x+y) &=48 \\\\ x-y &=\frac{48}{x+y} \\\\ \Rightarrow x-y &=\frac{48}{2 h+4}\;\;\;\;\;\;\;(3) \\\\ x+y&= 2 h+4\;\;\;\;\;\;\;\;(4) \end{aligned}\] Adding \((3) + (4)\), we get \[\begin{aligned} 2 x&=2 h+4+\frac{48}{2 h+4} \\\\ x&=h+2+\frac{24}{2 h+4}\;\;\;\;\;\;\;(5) \end{aligned}\] Using \((1)\) and \((5)\), we get \[(4 \sqrt{5})^{2}=h^{2}+\left(h+2+\frac{24}{2 h+4}\right)^{2}\] After expanding the previous expression, we get the equation \[h^{4}+6 h^{3}-14 h^{2}-96 h-32=0\] By a simple verification, we can show that \(-4\) and \(4\) are two roots of the equation and now let’s try to find the other roots \[\begin{aligned} h^{4}+6 h^{3}-14 h^{2}-96 h-32 &=(h-4)(h+4)\left(a h^{2}+b h+c\right) \\\\ &=\left(h^{2}-16\right)\left(a h^{2}+b h+c\right) \\\\ &=a h^{4}+b h^{3}+(c-16 a) h^{2}-16 b h-16 c \end{aligned}\] Therefore, \[\begin{aligned} &a=1 \\ &b=6 \\ &c=2 \end{aligned}\]