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Number Theory Total ordered pair in factorial equation

jacks

Well-known member
Apr 5, 2012
226
(A) Total no. of integer ordered pairs $(x,y)$ in $x^2-y! = 2001$

(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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
The smallest square greater than $2001$ is $45^2 = 2025 = 4! + 2001$. So that gives you one solution.

You have shown that $x$ is a multiple of $3$, and so $x^2$ is a multiple of $9$. But $2001$ is not a multiple of $9$. So if $x^2 - y! = 2001$ then $y!$ must not be a multiple of $9$. Therefore $y<6$. That severely limits the number of possible solutions!

For (B), $2013$ is also a multiple of $3$ but not a multiple of $9$. So you can apply a very similar argument in that case too.