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I try

$$2n-1| n^3 +1$$ $$\therefore 2n11 | 2n^3 + 2 -n^2(2n-1)$$ $$\therefore 2n-1| 4n-2$$

But but this is valid for all $n$. How to proceed? Thanks in advance

and I'm sorry if this is a duplicate, I don't see any similar questions

nonuser
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Helen
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  • In your line of $2n11 | 2n^3 + 2 -n^2(2n-1)$, I assume $2n11$ was meant to be $2n - 1$. However, from this line to the next, you say $2n-1| 4n-2$. I believe you made one or more calculation mistakes. Using your basic method, from the second line, you get $n^2 + 2$, so you can use $2n^2 + 4 - n(2n - 1) = n + 4$, then use $2n + 8$ and $2n + 8 - (2n - 1) = 9$, giving $2n - 1 | 9$ as in alex jordan's answer. – John Omielan Oct 06 '19 at 05:04
  • $!\bmod:! 2n!-!1!:\ n\equiv \frac{1}2\ $ so $\ 0\equiv n^3+1\equiv \frac{9}8\iff 9\equiv 0\iff 2n!-!1\mid 9\ \ $ – Bill Dubuque Oct 17 '23 at 05:31

4 Answers4

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Mod $(2n-1)$, you have that $2n\equiv1$. Now suppose $(2n-1)$ divides $n^3+1$. So always mod $(2n-1)$:

$$ \begin{align} n^3&\equiv-1\\ 8n^3&\equiv-8\\ (2n)^3&\equiv-8\\ (1)^3&\equiv-8\\ 1&\equiv-8\\ 9&\equiv0 \end{align} $$

So $2n-1$ must divide $9$. That doesn't leave too many options to check. (Be sure to look at both positive and negative divisors of $9$.)

2'5 9'2
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Hint. If $(2n-1)\mid(n^3+1)$, then at least $(2n-1)\mid8(n^3+1)$. Hence $$\frac{8(n^3+1)}{2n-1}=(4n^2+2n+1)+\frac9{2n-1}$$ is an integer. Hence $(2n-1)\mid 9$. Can you finish it now?

YiFan Tey
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Problem. Find all integers $n$ such that $2n - 1\, |\, n^3 + 1$.

Sol. Let $(a, b)$ denote the greatest common divisor (gcd) of $a$ and $b$. Then $$\begin{array}{lll} (2n - 1, n^3 + 1) &= (2n - 1, n^3 + 2n) \\ &= (2n - 1, n^2 + 2) \mbox{ since } (2n - 1, n) = 1 \\ &= (2n - 1, n^2 + 2 + 4n - 2) = (2n - 1, n(n + 4)) \\ &= (2n - 1, n + 4) \mbox{ since } (2n - 1, n) = 1 \\ &= (2n - 1 - 2(n + 4), n + 4) = (-9, n + 4). \end{array}$$ Since $2n - 1\, |\, n^3 + 1$, $2n - 1\, |\, 9$. So, $n$ may be $-4, -1, 0, 1, 2, 5$.

Yuanchuan Li
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$$ 2n-1\mid (2n-1)(4n^2+2n+1) = 8n^3-1$$

and $$2n-1\mid 8n^3+8$$ so $$2n-1\mid (8n^3+8) - (8n^3-1)=9$$

nonuser
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