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The problem is as such:

Find all natural numbers $n$ such that $$\dfrac{n^3+2021}{n+21}$$ is a natural number.

So far, I've only found two solutions. My approach: $$\begin{align}\frac{n^3+2021}{n+21}=\frac{(n+21)(n^2-21n+441)-7240}{n+21} &=n^2-21n+441-\frac{7240}{n+21}\end{align}$$

From this and the fact that $7240=2^3\times5\times181$, I got $n = 160$ and $n = 7219$. However, in the solution, it is stated that there are $9$ natural numbers $n$ that satisfy the expression. What am I missing here?

ryan.zcd
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  • $n+21$ can be equal to $\pm 2, \pm2^2,\pm 2^3$, $\pm 5$, and $\pm 181$, and other combination that can be cancelled out with numerator. Then you need to pick those that result in natural numbers. – Kafka Apr 03 '23 at 01:17
  • note that $5\times 181 = 905$, then $884$ is another solution, and with the other combinations you obtain the rest of the solutions – ZAF Apr 03 '23 at 01:17
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    Find all divisors of $7240$ greater than $21$ and subtract $21$ from them. – Geoffrey Trang Apr 03 '23 at 01:19

1 Answers1

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The nine solutions are:

$n_{1} = 181 - 21$

$n_{2} = 181 \times 5 - 21$

$n_{3} = 181 \times 2 - 21$

$n_{4} = 181 \times 2^{2} - 21$

$n_{5} = 181 \times 2^{3} - 21$

$n_{6} = 181 \times 2 \times 5 - 21$

$n_{7} = 181 \times 2^{2} \times 5 -21$

$n_{8} = 181 \times 2^{3} \times 5 - 21$

$n_{9} = 2^{3} \times 5 -21 $

ZAF
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