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

Question

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.

Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. — José Carlos Santos, Nov 08 '20 at 10:50
You have a sign typo: It should be $,12n+5,\ 5n+2,,$ or $,12n-5,\ 5n-2,$ in order for them to be relatively prime. The same methods in the linked dupes work for such. — Bill Dubuque, Nov 08 '20 at 11:58

Arthur · Accepted Answer · 2020-11-08T11:03:45.487

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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 $$ 12n+5=-7\\ 5n-2=-7 $$

edited Nov 08 '20 at 11:03

answered Nov 08 '20 at 10:53

Arthur

199,419

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Why are people downvoting this? It's a good answer. – Luke Collins Nov 08 '20 at 11:02
Upvoted, I also have no idea why this got downvoted. – Anonymous Nov 08 '20 at 11:05

Lion Heart · Answer 2 · 2020-11-08T11:04:58.277

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

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

edited Nov 08 '20 at 11:04

answered Nov 08 '20 at 10:58

Lion Heart

7,073

What if $n=20$? – Luke Collins Nov 08 '20 at 10:59

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

2 Answers2