shobber wrote: Find all positive integers $n$ such that $\dfrac{n^3+3}{n^2+7}$ is a positive integer! $\dfrac{n^3+3}{n^2+7} =n+\dfrac{3-7n}{n^2+7}$ Since it is an integer we have $\mid 3-7n \mid\geq n^2+7$.....

As socrates said, $\dfrac{n^3+3}{n^2+7} = n + \dfrac{3-7n}{n^2+7}$ We examine $f: N\rightarrow N : x \rightarrow n^2+7$ and $g: N\rightarrow N : x\rightarrow 3-7n$ When $x=0$, $f(0) > g(0)$ We search the intersection of the two graphs: $n^2+7n+4=0$ There are no solutions of n in this equation, so the two graphs never intersect. \[ \Rightarrow n^2+7 > 3-7n \] So $\dfrac{3-7n}{n^2+7} \notin N$ There exist no positive integers $n$ such that $\dfrac{n^3+3}{n^2+7}$ is an integer.

Jan I'm afraid you are not right. Just take n=2.... What I said is to solve the inequation $\mid 3-7n \mid\geq n^2+7$ Do you get it?? If not tell me again about it

Oh, I'm sorry... It's wrong because I said that the equation $n^2+7n+4=0$ has no integer solutions for $n$. (I had to be quick because my mom already called me twice to come and eat. ) So factoring $n^2+7n+4$ gives $(x-x_1)(x-x_2)$ Between these two points, $3-7n \geq n^2 + 7$, We notice that $-1<x_1<x_2<0$ so there are no integer values of n between them. If $3-7n < 0$ , we get the equation $3-7n \leq -n^2-7$ Working out gives: $n \in [2,5]$ Bruteforcing tells us that $2$ and $5$ are the only possibilities: $\dfrac{2^3+3}{4+7}=1$ and $\dfrac{5^3+3}{32}=4$ Socrates, that's your solution, only in a long and less elegant way. I'm just posting this to show that my solution wasn't completely useless (if I only calculated the quadratic correctly... )

shobber wrote: Find all positive integers $n$ such that: \[ \dfrac{n^3+3}{n^2+7} \]is a positive integer. Note that: $$\frac{n^3+3}{n^2+7}=\frac{n^3+7n-7n+3}{n^2+7}=n+\frac{3-7n}{n^2+7}$$Now since $\frac{n^3+3}{n^2+7}$ is an integer, hence $\frac{3-7n}{n^2+7}$ is an integer: That means that $3-7n \ge n^2+7$ or $7n-3 \ge n^2+7$. Now if $3-7n \ge n^2+7$. $$n^2+7n+4 \le 0 \implies \; \text{no such} \; n$$This is becuase $n \in \mathbb N$ implies $n \ge 1$ and then at least $12 \le 0$, contradiction!. Hence $7n-3 \ge n^2+7$ implies: $$n^2-7n+10 \le 0 \implies (n-2)(n-5) \le 0 \implies n \in (2, 5)$$By some calculations you get that $n=2$ or $n=5$. Thus we are done