11Here's the outline
If y>4, then 3|x2 implies 9|x2 so if y>5, 9 must divide 2001, which is a contradiction.
\hspace{-16}$Find all ordered pairs $\bf{(x,y)}$ in $\bf{x^2-y!=2001}$\\\\ Where $\bf{x,y\in \mathbb{N}}$
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2 Answers
1057x=45 and y=4 is a solution....
and if y is greater than 4 then atleast unit' digit of y is 0....so x should have unit's digit as either 1 or 9....i guess sumthing cud be worked out like dat...
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