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Let $ x,~y~$ are two natural numbers such that $~x,~y \in (0,1001)~$ and also satisfying $~x^2=1+4y.$ Then how many ordered pair of such $~(x,y)~ $ are possible?

First notice that , $x$ is odd natural number, so there exists a positive integer $n$ such that $~x=2n+1.$ Then we have $~y=n^2+n.$

From here how can I find the possible pairs of $~(x,y)~$ such that $~x,y~$ are natural numbers and $x^2=1+4y.$

Please help me to solve this.

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  • $\begingroup$ What does $x,y \in (0,1001)$ have to do with the rest of the question? How are $x,y$ related to $p,q$? $\endgroup$
    – Erick Wong
    Jun 6, 2021 at 7:21
  • $\begingroup$ @Erick Thank you. They are not $p, q$ instead they are $x,y$. $\endgroup$
    – abcdmath
    Jun 6, 2021 at 8:41

1 Answer 1

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$$ q = \frac{p^2 - 1}{4} < 1001 \implies p^2 < 4005 \implies p \le 63. $$ $$ p = 2n+1 \le 63 \implies n \le 31. $$ It's easy to see that all pairs $\big(2n+1,n(n+1)\big)$ satisfy the equation and constrains for $\,n = 1, \ldots 31$.

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