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Question

Prove that the fraction m(n+1)+1m(n+1)n is irreducible for every positive integers m and n.

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Solution

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+2p2q 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)2q1=4pq+4p+14pq+4p2q1 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)2q1=4pq+4p+2q+2+14pq+4p+2q+22q1
=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.

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