Prove that for any integer $n$, the fraction $\frac{3n+2}{4n+3}$ is reduced.

I need help with this question. Prove that for any integer $n$, the fraction $\frac{3n+2}{4n+3}$ is reduced.

Solution 1:

Note that $$3(4n+3)-4(3n+2)=1.$$ So any integer that divides both $4n+3$ and $3n+2$ must divide $1$. And nothing $\gt 1$ divides $1$, so the greatest common divisor of $4n+3$ and $3n+2$ is $1$.

Solution 2:

The following proof repeatedly uses the fact that if $m$ divides $a$ and $m$ divides $b$, then $m$ divides $a+kb$ for any integer $k$.

$m$ is a common factor of $4n+3, 3n+2$

implies

$m$ is a common factor of $n+1, 3n+2$

implies

$m$ is a common factor of $n+1, -1$

implies

$m$ is a factor of $-1$

implies

$m=\pm 1$