Question

Prove that the fraction $\frac{m(n+1)+1}{m(n+1)-n}$ is irreducible for every positive integers $m$ and $n$.

Answer 1

Let us consider three cases as follows:

Case 1: When $m$ is even and $n$ is even, then $m=2p$ and $n=2q$. Therefore,

$\frac{m(n+1)+1}{m(n+1)-n}=\frac{2p(2q+1)+1}{2p(2q+1)-2q}=\frac{4pq+2p+1}{4pq+2p-2q}$ which is not reducible.

Case 2: When $m$ is even and $n$ is odd, then $m=2p$ and $n=2q+1$. Therefore,

$\frac{m(n+1)+1}{m(n+1)-n}=\frac{2p(2q+1+1)+1}{2p(2q+1+1)-(2q+1)}=\frac{2p(2q+2)+1}{2p(2q+2)-2q-1}=\frac{4pq+4p+1}{4pq+4p-2q-1}$ which is not reducible.

Case 3: When $m$ is odd and $n$ is odd, then $m=2p+1$ and $n=2q+1$. Therefore,

$\frac{m(n+1)+1}{m(n+1)-n}=\frac{(2p+1)(2q+1+1)+1}{(2p+1)(2q+1+1)-(2q+1)}=\frac{(2p+1)(2q+2)+1}{(2p+1)(2q+2)-2q-1}=\frac{4pq+4p+2q+2+1}{4pq+4p+2q+2-2q-1}$

$=\frac{4pq+4p+2q+3}{4pq+4q+1}$ which is not reducible.

Hence the fraction $\frac{m(n+1)+1}{m(n+1)-n}$ is irreducible for every positive integer $m$ and $n$.