If an integer $a$ is coprime with an integer $b$, then will the integer $a$ also be coprime with the integer $9a + b$?

Question

If an integer $a$ is coprime with an integer $b$, then will the integer $a$ also be coprime with the integer $9a + b$ ?

Notice that if $d|a$ and $d|(9a+b)$, then $d|b$ since $b=(9a+b)-9a$; so $d=1$ if $d\in N$. — user84413, Aug 27 '14 at 00:43

score 4 · Answer 1 · answered Aug 27 '14 at 00:40

4

If $1=ma+nb$, then $1=(m-9n)a+n(9a+b)$.

answered Aug 27 '14 at 00:40

Kim Jong Un

See http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity for context. – blue Aug 27 '14 at 00:46
Using Bezout here is a overkill, since it can be done simply by divisibility. Indeed, it is true in any ring (not only rings satisfying Bezout's identity). – Bill Dubuque Aug 27 '14 at 01:01

score 0 · Answer 2 · edited Apr 13 '17 at 12:19

0

Hint $\ $ The linear map $\,(a,b)\to (a,9a\!+\!b)\,$ has det $=\color{#c00}1,\ $ hence $\ \gcd(a,9a\!+\!b)\mid \color{#c00}1\cdot \gcd(a,b)$

edited Apr 13 '17 at 12:19

Community

answered Aug 27 '14 at 01:12

Bill Dubuque

2 Answers2