An irreducible fraction

Prove that the fraction \dfrac{21n + 4}{14n + 3} is irreducible for every natural number n.

A common divisor of two numbers divides every integer combination of them.

Find integers a, b with a(14n+3) + b(21n+4) not depending on n.

Solution. Let d = \gcd(21n+4,\, 14n+3). Then d divides 3(14n+3) - 2(21n+4) = 42n + 9 - 42n - 8 = 1, so d = 1 and the fraction is already in lowest terms.

This was the very first problem of the very first International Mathematical Olympiad, held in Romania in 1959.