MathLabs

Problem 1

Prove that the fraction 21n+414n+3\frac{21n+4}{14n+3} is irreducible for every natural number nn.
Step 2 of 4: Find a Bézout identity
In plain words

A linear combination of the numerator and denominator equal to 11 forces their gcd to divide 11.

3(14n+3)−2(21n+4)=13(14n+3)-2(21n+4)=1
Detailed analysis

A direct calculation gives 3(14n+3)−2(21n+4)=42n+9−42n−8=13(14n+3)-2(21n+4)=42n+9-42n-8=1. Thus every common divisor of the two terms divides 11.