Prove that the fraction m(n+1)+1m(n+1)−n is irreducible for every positive integers m and n.
Let us consider three cases as follows:
Case 1: When m is even and n is even, then m=2p and n=2q. Therefore,
m(n+1)+1m(n+1)−n=2p(2q+1)+12p(2q+1)−2q=4pq+2p+14pq+2p−2q which is not reducible.
Case 2: When m is even and n is odd, then m=2p and n=2q+1. Therefore,
m(n+1)+1m(n+1)−n=2p(2q+1+1)+12p(2q+1+1)−(2q+1)=2p(2q+2)+12p(2q+2)−2q−1=4pq+4p+14pq+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,
m(n+1)+1m(n+1)−n=(2p+1)(2q+1+1)+1(2p+1)(2q+1+1)−(2q+1)=(2p+1)(2q+2)+1(2p+1)(2q+2)−2q−1=4pq+4p+2q+2+14pq+4p+2q+2−2q−1
=4pq+4p+2q+34pq+4q+1 which is not reducible.
Hence the fraction m(n+1)+1m(n+1)−n is irreducible for every positive integer m and n.