Show that $12n+5$ and $5n-2$ are relatively prime for all $n$ (in $\mathbb{Z}$)

Question

Show that $12n+5$ and $5n-2$ are relatively prime for all $n$ (in $\mathbb{Z}$)

I've seen similar questions on here, I tried following their answers and doing it myself, but it's not working out for me. I tried the Euclidean Algo a few times and it's never ending. I'd be happy if someone could post the way to do this.

Answer 1

The Euclidean algorithm might not reach $0$, but it will help greatly: $$ \begin{array}{cc} 12n+5&5n-2\\ 12n+5-2(5n-2)&5n-2\\ 2n+9&5n-2\\ 2n+9&5n-2-2(2n+9)\\ 2n+9&n-20\\ 2n+9-2(n-20)&n-20\\ 49&n-20 \end{array} $$ So basically, if $n-20$ has a factor of $7$, then $12n+5$ and $5n-2$ will both have a factor of $7$ (and similarly for a factor of $49$). For instance, for $n=-1$ we get $$ 12n+5=-7\\ 5n-2=-7 $$

Answer 2

$$(12n+5,5n-2)=(2n+9,5n-2)=(2n+9,n-20)=(n+29,n-20)=(n+29,49)$$

for $n=7t+6$ where $t\in Z$ both can divisible by 7. They are not relatively prime.