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Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.

Attempt: We have \begin{equation*} \frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}. \end{equation*} So, we must have $(2n+1) \mid (n^2-4)$, so $n^2-4 = k(2n+1)$, for some $k \in \Bbb Z$. But, I did not be able to find $n$ from here. Any ideas? Thanks in advanced.

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    $\begingroup$ Hint : $$\frac{3n^2+4n+5}{2n+1}=\frac{1}{4}(6n+5+\frac{15}{2n+1})$$ $\endgroup$
    – Peter
    Apr 10 at 11:56
  • $\begingroup$ Try to use Polynomial Long Division $\endgroup$ Apr 10 at 11:58
  • $\begingroup$ @Peter Thanks, Sir! $\endgroup$
    – gerrr
    Apr 10 at 13:51

3 Answers 3

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First way: Using the Extended Euclidean Algorithm in $\Bbb Q[x]$.

Notice that \begin{equation*} 15 = 4(3n^2+4n+5) - (6n+5)(2n+1). \end{equation*} Hence, if $2n+1$ divides $3n^2+4n+5$, then it also divides $15$. Thus, $2n+1 \in \{\pm 1, \pm 3, \pm 5, \pm 15\}$, i.e., \begin{equation*} n \in \{-8,-3,-2,-1,0,1,2,7\}. \end{equation*}

Second way: Just using the Elementary number theory.

From your approach, we have \begin{equation*} \frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}. \end{equation*} Now, let $k=2n+1$, then $2n \equiv -1 \pmod{k}. \ldots (1)$

We want $(2n+1) \mid (n^2-4)$, i.e., $n^2-4 \equiv 0 \pmod{k}. \ldots (2)$ Multiplying both side in $(2)$ by $4$, we have \begin{align*} 4n^2 - 16 \equiv 0 \pmod{k} &\Leftrightarrow (2n)^2 - 16 \equiv 0 \pmod{k} \\ &\Leftrightarrow (-1)^2 - 16 \equiv 0 \pmod{k} \qquad \qquad (\text{by $(1)$}) \\ &\Leftrightarrow -15 \equiv 0 \pmod{k}. \end{align*} Hence, $2n+1 \mid 15$. Thus, $2n+1 \in \{\pm 1, \pm 3, \pm 5, \pm 15\}$, i.e., \begin{equation*} n \in \{-8,-3,-2,-1,0,1,2,7\}. \end{equation*}

Now, we are done.

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Another approach:

By direct division we find:

$3n^2+4n+5=(\frac 32 n+\frac 54)(2n+1)+\frac{15}4$

multiplying both sides by $4$ and simplify we get:

$12n^2+16n+5=(6n+5)(2n+1)$

Or:

$\frac{4(3n^2+4n+5)}{2n+1}=6n+5+\frac {15}{2n+1}$

Therefore $2n+1$ must divide $15$ that is it must be equal to one of its divisors which are $\pm 1$, $\pm 3$, $\pm 5$ and $\pm 15$

1): $2n+1=1\rightarrow n=0$, $2n+1=-1\rightarrow n=-1$

2): $2n+1=3\rightarrow n=1$, $2n+1=-3 \rightarrow n=-2$

3): $2n+1=5\rightarrow n=2$, $2n+1=-5 \rightarrow n=-3$

4): $2n+1=15 \rightarrow n=7$, $2n+1=-15 \rightarrow n=-8$ so solutions are:

$0, \pm 1, \pm 2, -3, 7, -8$

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Write $k=2n+1$ then $n=(k-1)/2$ so $$3n^2+4n+5= {3(k^2-2k+1) + 8(k-1)+20\over 4} ={3k^2+2k+15\over 4}$$ and thus $$4k\mid 3k^2+2k+15\implies k\mid 15$$

Now you have only few values of $k$...

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