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I want to be able to take a function and find all the ordered pairs where both the $x$ coordinate and $y$ coordinate are natural numbers. I have been working with the equation $$f(x) =\frac{-x^2+45x}{2x+1}$$ which I know has a finite number of natural numbered ordered pairs, ((3,18) and (6,18)) but I have no idea how to find them without using guess and check. Is guess and check the only option or is there a better way to find them?

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  • $\begingroup$ Well, you can play with this expression a bit. Easy to see that $\gcd (x,2x+1)=0$ so you want $2x+1$ to divide $45-x$. But if this holds then $2x+1$ divides $90-2x=91-(1+2x)$ so... $\endgroup$
    – lulu
    Commented Jul 18, 2017 at 18:31
  • $\begingroup$ wouldn't the $\operatorname{gcd}(x,2x+1)=1$ $\endgroup$ Commented Jul 18, 2017 at 19:35

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The standard procedure in such problems is to try to find a function of degree lower than the denominator which the denominator divides. In this case, since the denominator is of degree 1, we seek a constant term which it divides.

Notice that $-x^2+45 = x(45-x)$. $\gcd(x, 2x+1) = 1$, so $2x+1$ has to divide $45-x$. At this stage, we need a trick. Here, we spot that $2x+1 \mid 45-x$ implies that $2x+1 \mid 2(45-x)$ $ = 90 - 2x = $ $ 91 - (2x+1)$. Hence, $2x+1 \mid 91$.

Now, all you need to do is check the factors of $91 ( = 7 \times 13)$. This yields $x = 0, 3, 6, 45$. However, $x,y$ are natural numbers, so $x = 0$ is invalid; and $x=45$ yields $y=0$, so this case is also invalid. Checking the other two we see that they are valid.

Therefore, $x = 3$ or $6$.

(Note that usually you might have to consider the negative factors of $91$, but here one can clearly see that this would not yield any solutions $\in \mathbb{N}$)

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  • $\begingroup$ Thank you! Is there a numerical method though or is this the only way? $\endgroup$
    – Faraday
    Commented Jul 19, 2017 at 1:06
  • $\begingroup$ @Frank Not that I can think of off the top of my head. $\endgroup$
    – Plato
    Commented Jul 19, 2017 at 7:37

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