Xavier Aptitude Test 2021 Paper

Since ${\mathit{x}}^2-{\mathit{y}}^2=20$ and $\mathit{x},\mathit{y},\mathit{z}$ are positive integers,
$\left(\mathit{x}+\mathit{y}\right)*\left(\mathit{x}-\mathit{y}\right)=20$, Hence $\mathit{x}-\mathit{y},\mathit{x}+\mathit{y}$ are factors of $20$.
Since $\mathit{x},\mathit{y}$ are positive integers, $\mathit{x}+\mathit{y}$ is always positive, and for the product of $\left(\mathit{x}+\mathit{y}\right)*\left(\mathit{x}-\mathit{y}\right)$ to be positive $\mathit{x}-\mathit{y}$ must be positive.
$\mathit{x},\mathit{y}$ are positive integers and $\mathit{x}-\mathit{y}$ is positive $\mathit{x}$ must be greater than $\mathit{y}$.
The possible cases are : $\left(\mathit{x}+\mathit{y}=10,\mathit{x}-\mathit{y}=2\right),\left(\mathit{x}+\mathit{y}=5,\mathit{x}-\mathit{y}=4\right)$
The second case fails because we get $\mathit{x}=\frac{9}{2},\mathit{y}=\frac{1}{2}$ but $\mathit{x},\mathit{y}$ are integral values
For case one $\mathit{x}=6,\mathit{y}=4$
${\mathit{y}}^3-2{\mathit{x}}^2-4\mathit{z}\ge -12$
Substituting the values of $\mathit{x}$ and $\mathit{y}$, we have :
$64-72-4\times \mathit{z}\ge -21$
$-8-4\times \mathit{z}\ge -12$
$\mathit{z}\le 1$
Since $\mathit{x},\mathit{y},\mathit{z}$ are positive integers, the only possible value for $\mathit{z}$ is $1$.