Prove `(21n+4)/(14n+3)` Is Irreducible For Every Natural Number `n`

Prove `(21n+4)/(14n+3)` Is Irreducible For Every Natural Number `n`. 

Solution

 Solution 01

Denoting the greatest common divisor (GCD) of `a` and `b` `(a,b)` and we will use Euclidean Algorithm Theory.

                `(21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1`

It follows that `(21n+4)/(14n+3)` is irreducible. Q.E.D

 Solution 02

In this method, we will prove by Contradiction Premise.

Assuming that:     `(21n+4)/(14n+3)` is a reducible fraction where `p` is satisfied that:

                `(21n+4)\equiv0(modp)`

Then,         `(42n+8)\equiv0(modp)`

Subtracting the second equation:  `[3(14n+3)-1]\equiv0(modp)`

We clearly see that:     `1\equiv0(modp)` Which is clearly absurd.

Hence, `(21n+4)/(14n+3)` is irreducible. Q.E.D

 Solution 03

This method also use the Contradiction.

Assuming that:     `(21n+4)/(14n+3)` is a reducible fraction.

If a certain fraction `a/b` is reducible, then the fraction `(2a)/(3b)` must be reducible.

In this case,     `(2a)/(3b)=(42n+8)/(48n+9)`

This fraction consists of two consecutive numbers, which never share any factor.

So in this case, `(2a)/(3b)` is irreducible fraction, which is absurd.

Hence, `(21n+4)/(14n+3)` is irreducible. Q.E.D

Solution by: Thin Sokkean