1959 IMO Problems/Problem 1

Problem

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


Solutions

First Solution

We observe that

$\displaystyle 3(14n+3) = 2(21n+4) + 1.$


Since a multiple of $\displaystyle 14n+3$ differs from a multiple of $\displaystyle 21n+4$ by 1, we cannot have any postive integer greater than 1 simultaneously divide $\displaystyle 14n+3$ and $\displaystyle 21n+4$. Hence the greatest common divisor of the fraction's numerator and denominator is 1, so the fraction is irreducible. Q.E.D.

Second Solution

Denoting the greatest common divisor of $\displaystyle a, b$ as $\displaystyle (a,b)$, we use the Euclidean algorithm as follows:

$\displaystyle ( 21n+4, 14n+3 ) = ( 7n+1, 14n+3 ) = ( 7n+1, 1 ) = 1$

As in the first solution, it follows that $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D.

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