Find the $\gcd(2k-1,9k+4)$

Question

This is that I did so far.If $d=\gcd(2k-1,9k+4)$ then:

$d \mid 2k-1$ and $d \mid 9k+4$ so $d \mid 2(9k+4)-9(2k-1)\implies d\mid 17$ so $d$ is $17$ for some values and $1$ for others.

After some observation I concluded that if $k=9+17n$ then $d=17$ and $d=1$ otherwise but I can’t prove it. Any help will be appreciated

So you want to find which $k$ satisfy $17 \mid 2k-1$.In particular, you observe that this is true if and only if $ 17 \mid k - 9$. How can you work with that? — Calvin Lin, Apr 03 '23 at 23:32
Hint : Try to find a solution to this$$d\equiv-1[2];and;d\equiv4[9]$$ — Moustapha_M_I, Apr 03 '23 at 23:48
Easier and (more general) to use the slight generization of the Euclidean algorithm explained in the linked dupes, yielding gcd $= (2k-1,17).,$ It's $> 1!\iff! \bmod 17!:\ 2k\equiv 1\equiv 18!\iff! k\equiv 9\ \ $ — Bill Dubuque, Apr 04 '23 at 00:15

score 1 · Answer 1 · answered Apr 03 '23 at 23:31

1

Do the Euclidean algorithm: add/subtract copies of one slot of $\gcd(-,-)$ from the other.

$$ \gcd(2k-1,9k+4) = \gcd(2k-1, 9k+4 - 4(2k-1)) = \gcd(2k-1, k+8) = \gcd(2k-1-2(k+8), k+8) = \gcd(-17, k+8). $$

answered Apr 03 '23 at 23:31

Dennis

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 04 '23 at 00:19
Understood, I’ll try to avoid in the future. Thanks! – Dennis Apr 04 '23 at 01:28

1 Answers1