Prove that \(\frac{21n+4}{14n+3}\) is irreducible for every natural number \(n\).

Solution

Method 01: Let \((a,b)\) denote the greatest common divisor of \(a\) and \(b\). Using the Euclidean Algorithm,

\[ (21n+4,14n+3)=(7n+1,14n+3)=(7n+1,1)=1. \]

Therefore, \(\frac{21n+4}{14n+3}\) is irreducible. Q.E.D.

Method 02: We prove by contradiction. Assume that \(\frac{21n+4}{14n+3}\) is reducible. Then there exists a prime number \(p\) such that \(21n+4\equiv0\pmod p\) and \(14n+3\equiv0\pmod p\).

Multiplying the first equation by \(2\), we get \(42n+8\equiv0\pmod p\). Subtracting \(3(14n+3)\equiv0\pmod p\),

\[ (42n+8)-(42n+9)\equiv0\pmod p \]

so \(-1\equiv0\pmod p\), which is impossible. Hence, \(\frac{21n+4}{14n+3}\) is irreducible. Q.E.D.

Solution by: Thin Sokkean