Consecutive Integers?

Suppose that $m > 2n - 1$ .Then,

$(2n - 1)^3 < 8n^3 - 10n^2 - n + 2 \\ \iff 2n^2 - 7n +3 > 0 \\ \iff (2n - 1)(n - 3) > 0 \\ \iff n < \frac{1}{2}, n > 3$ .

Also suppose that $m < 2n$ . Then,

$(2n)^3 > 8n^3 - 10n^2 - n + 2 \\ \iff 10n^2 + n -2 > 0 \\ \iff (5n - 2)(2n + 1) > 0 \\ \iff n < -\frac{1}{2}, n > \frac{2}{5}.$

Therefore $2n -1 < m < 2n$ and so $m$ cannot be an integer; unless $\frac{1}{2} \leq n \leq 3$ and $n < -\frac{1}{2} \text{ or } n > \frac{2}{5}$ , which gives $n = 1, 2, 3$ , giving the pairs $(-1, 1)$ and $(5, 3)$ ( $n = 2$ does not give an integer solution).

The only other possibility is when $n < \frac{1}{2} \text{ or } n > 3$ and $-\frac{1}{2} \leq n \leq \frac{2}{5}$ , which gives only $n = 0$ which yields no integer solutions for $m$ .

Thus the only pairs of solutions are $\boxed{(-1, 1) \text{ and } (5, 3)}$ .

Two solutions are (-1,1) and (5,3)

First one i have found by putting m= -n

Second one by Hit trial method

Can anyone give a proper wayout