Verify my proof of divisibility of a rational fraction I tried to prove that the following rational fraction can be divided only when n=1,2,5 without using mathematical induction or function, just by using basic method. $$\frac{n^2+2}{2n-1}$$  this expression can be divided only when the remainder of $$\frac{n^2}{2n-1}$$ is $$\frac{2n-3}{2n-1}$$ and the quotient of the following expression $$\frac{n^2}{2n-1}$$ has the same quotient with the following expression, except in the case of n=1$$\frac{n^2}{2n}$$ because when some dividend is divided by these two different divisor, in order to $$2n-1$$ has much more quotient than $$2n$$ It's necessary to meet particular condition that when dividend is greater than or equal to $$2n$$ also when the dividend is greater than or equal to $$(2n)(2n-1)$$ then $$2n-1$$ can has much more quotient than $$2n$$ and when the dividend is less than $$2n$$ also when the value of the dividend is equal to the value of $$2n-1$$ then $$2n-1$$ can has much more quotient than $$2n$$ So $$n^2$$ is greater than or equal to $$2n$$ except when n=1, and when n=1, the value of $$n^2$$ is less than $$2n$$ and is equal to the value of $$2n-1$$ So when n=1, $$2n-1$$ has much more quotient than $$2n$$ and when n>1, $$n^2$$ is greater than or equal to $$2n$$ and is less than $$(2n)(2n-1)$$ So $$\frac{n^2}{2n-1}$$ has the same quotient with $$\frac{n^2}{2n}$$ except in the case of n=1. and $$\frac{n^2}{2n}=\frac{n}{2}$$ so the quotient of $$\frac{n}{2}$$ can be expressed by a formula $$\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}$$ and in the following expression$$\frac{n}{2}$$ when n is even number there is no remainder and when n is odd number the remainder is always $$\frac{1}{2}$$ It also can be expressed as $$\frac{n}{2n}$$ and in the following expression $$\frac{n^2}{2n-1}$$ when n is even number the remainder of the expression is as much as Its quotient. and when n is odd number the remainder of the division is $$n+quotient$$ now I can answer the question about when the remainder become $$2n-3$$ is, by using the formula $$\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}$$ when n is even number It can be expressed as $$\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}=2n-3$$ this equation can be changed into $$12-(1+(-1)^{n+1})=6n$$ by multiplying both sides by 4 and by transposition. and when n is odd number, It can be expressed as $$n+\frac{n}{2}-\frac{1+(-1)^{n+1}}{4}=2n-3$$ this equation can be changed into $$12-(1+(-1)^{n+1})=2n$$ so, the following expression $$\frac{n^2+2}{2n-1}$$ is can be divided only when n=2,5 but this equation works only when n>1. and when=1 the expression can be divided. so this is what I tried to prove the question. I wanna know this proof is correct or incorrect, Could you please check my proof? roughly If you want. and I wanna know this kind of approach is proper or not when I'm gonna study advanced mathematics.
 A: I would suggest the following approach :
First consider $(2n-1)(2n+1)=4n^2-1$ hence we have $$4\cdot \frac{n^2+2}{2n-1}=\frac{4n^2+8}{2n-1}=\frac{4n^2-1}{2n-1}+\frac{9}{2n-1}=2n+1+\frac{9}{2n-1}$$
Since $2n-1$ is odd, it must divide $n^2+2$ , if it divides $4(n^2+2)$ (try to find out why), hence you only have to check for which $n$ we have $2n-1|9$. Try to solve this.
A: By long division,
$$\frac{n^2+2}{2n-1}=\frac n2+\frac14+\frac9{4(2n-1)}$$ or
$$4\frac{n^2+2}{2n-1}=2n+1+\frac9{2n-1}$$
so that $2n-1$ must divide $9$ and $n$ could be $1,2$ or $5$, which turn out to all be valid solutions.
If negatives are allowed, consider $0,-1,-4$ which are also possible.
A: Yet another way to get the answer:
If $2n - 1$ divides $n^2 + 2$ then there is an integer $k$ such that
$n^2 + 2 = k(2n - 1),$
that is, 
$$ n^2 - 2kn + (k + 2) = 0.$$
Solving this as a quadratic equation in $n,$
\begin{align}
n &= \frac{2k \pm \sqrt{4k^2 - 4(k + 2)}}{2} \\
 &= k \pm \sqrt{k^2 - k - 2)},
\end{align}
which is possible only if $k^2 - k - 2 = m^2$ 
for some non-negative integer $m.$
Observing that $k^2 - k - 2 = (k - 2)(k + 1),$
the product can be a square only if it is non-negative, that is,
$k \geq 2$ or $k \leq = -1.$
If $k \geq 2$ then $m \leq k - 1$; it follows that 
$m^2 \leq k^2 - 2k + 1,$ and therefore
$$k^2 - k - 2 \leq k^2 - 2k + 1.$$
Canceling the $k^2$ terms and collecting the other terms appropriately, this is equivalent to $k \leq 3.$ We can check that $k = 2$ and $k = 3$ both are valid solutions.
If $k \leq -1,$ let $p = -k$; then $p \geq 1$ and $m^2 = p^2 + p - 2,$
which has solution $m = 0$ if $p = 1,$ solution $m = 2$ if $p = 2,$
and otherwise $m \geq p + 1,$ so $m^2 \geq p^2 + 2p + 1,$
so $p^2 + p - 2 \geq p^2 + 2p + 1,$ so $-p \geq 3,$ but that contradicts the case assumption that $k \leq -1.$
Hence there are no other solutions.
So the solutions are $k = 2,$ $k = 3,$ $k = -1,$ and $k = -2$.
The possible values of $n$ therefore are given in this table:
\begin{array}{ccccl}
 k  &\qquad k^2 - k - 2 \qquad&  &n& \\ \hline
\phantom{-}2  &  4 - 2 - 2 = 0  &  2 \pm \sqrt0 &=& 2, \\
\phantom{-}3  &  9 - 3 - 2 = 4  &  3 \pm \sqrt4 &=& 1 \text{ or } 5, \\
-1  &  1 - (-1) - 2 = 0  &  -1 \pm \sqrt0 &=& -1, \\
-2  &  4 - (-2) - 2 = 4  &  -2 \pm \sqrt4 &=& 0 \text{ or } -4.
\end{array}

It's easier to solve $2n - 1 \mid 9,$ so I would go with one of the other solutions.
