[1] Total no. of Integer ordered pairs $(x,y)$ for which $x^2-y! = 2001$
[2] Total no. of Integer ordered pairs $(x,y)$ for which $x^2-y! = 2013$
My Try:: (1) $x^2-y! = 2001\Rightarrow x^2 = 2001+y!$
We Know that $y!$ end with $0, 1,2,4,6$ and last digit of $x^2$ is $0,1,4,5,6,9$
But I Did Not understand How can I proceed further
Help Required,
Thanks
On
As Calvin answered the first one, here is a hint for the other:
$$2013 \equiv 5 \pmod 8$$
If $x$ is odd, then $x^2 \equiv 1 \pmod 8$.
When $y \geq \cdots$ you know $8 \mid y!$.
On
We'll check $\left[1\right]$.
$x^{2} \geq 2002\quad$. Then, $x > 44.\quad$ Let's $x = 44 + \delta.\quad$. $\delta > 0.\quad$ Then
$$ \delta^{2} + 88\delta + y! - 65 = 0\,, \quad \delta_{\pm} = -44 \pm \sqrt{2001 - y!} < 0\ \mbox{for any}\quad y\ \ni y! \leq 2001 $$
$\color{#ff0000}{\large\mbox{Then,}\quad\color{#0000ff}{\left[1\right]\quad\mbox{is false.}}}$
Hint: 2001 (and 2013) is a multiple of 3, but not 9.
Hint: There are no squares that are multiples of 3 but not 9.
Hint: There are finitely many factorials that are multiples of 3, but not 9.
Hence, you only need to check finitely many cases. In particular, you only need to check when $y = $(fill in the blank).