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(B) Total no. of integer ordered pairs $(x,y)$ in $x^2-y! = 2013$

My Trail :: (A) Given $x^2-y! = 2001 = 3 \times 23 \times 29 \Rightarrow x^2 -y! = \left(3\times 23 \times 29\right)$

means $(x^2-y!)$ must be divisible by $3$ So $x = 3k$ anf $y\geq 3$, where $k\in \mathbb{Z}$

So $9k^2-y! = 3\times 23 \times 29$

Now How can i solve after that

Thanks