Xavier Aptitude Test 2023 Paper

The given equation is,
$\mathrm{x}\left(\mathrm{x}-\mathrm{p}\right)-\mathrm{y}\left(\mathrm{y}+\mathrm{p}\right)=7\mathrm{p}$
${\mathit{x}}^2-\mathit{p}\mathit{x}-{\mathit{y}}^2-\mathit{p}\mathit{y}=7\mathit{p}$
${\mathit{x}}^2-{\mathit{y}}^2-\mathit{p}\mathit{x}-\mathit{p}\mathit{y}=7\mathit{p}$
$\left(\mathit{x}+\mathit{y}\right)\left(\mathit{x}-\mathit{y}\right)-\mathit{p}\left(\mathit{x}+\mathit{y}\right)=7\mathit{p}$
$\left(\mathit{x}-\mathit{y}-\mathit{p}\right)\left(\mathit{x}+\mathit{y}\right)=7\mathit{p}$
As '7' & 'p' both are prime numbers
$\left(\mathit{x}-\mathit{y}-\mathit{p}\right)\left(\mathit{x}+\mathit{y}\right)$ can be expressed as $\left(7\times \mathit{p}\right)$ or $\left(7\mathit{p}\times 1\right)$
Case (i) - $\left(\mathit{x}+\mathit{y}\right)\times \left(\mathit{x}-\mathit{y}-\mathit{p}\right)=7\times \mathit{p}$
$\mathit{x}+\mathit{y}+\mathit{x}-\mathit{y}-\mathit{p}=7+\mathit{p}$
$2\mathit{x}-\mathit{p}=7+\mathit{p}$
$\mathit{x}=\frac{7}{2}+\mathit{p}$

But it's given that ' $\mathit{x}$ ' is a positive integer. This case is not possible.
Case (ii) - $\left(\mathit{x}+\mathit{y}\right)\times \left(\mathit{x}-\mathit{y}-\mathit{p}\right)=7\mathit{p}\times 1$
$\mathit{x}+\mathit{y}+\mathit{x}-\mathit{y}-\mathit{p}=7\mathit{p}+1$
$2\mathit{x}-\mathit{p}=7\mathit{p}+1$
$\mathit{x}=\frac{1}{2}+4\mathit{p}$

But it's given that ' $\mathit{x}$ ' is a positive integer. This case is not possible.
The given equation is not possible with given conditions.
Option (E) is correct.