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Show: $\gcd(9k+4,2k+1)=1 ~~ \forall k\in \mathbb Z$

Indirect proof.

If  $1\neq d=\gcd(9k+4,2k+1)~\exists k\in \mathbb Z$,
then $d$ has to be of the form $2m+1$ for an integer $m$.

That somehow throws me back at the beginning.

mle
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355durch113
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6 Answers6

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Hint $\ $ If $\,d\,$ is a common factor of both then $\,{\rm mod}\ d\!:\ 2k+1\equiv 0\equiv 9k+4.\,$ Now eliminate $\,k\,$ by cross multiplying, to obtain $\,0 \equiv \color{}9(2k+1)-\color{}2(9k+4)\equiv \color{#0a0}9-\color{#c00}8\equiv 1\,\Rightarrow\,d\mid 1,\,$ so $\,d = 1.$

Remark $\ $ More conceptually, $ $ put the fractions for $\,-k\,$ over the common denominator $\,18$

$$ {\rm mod}\ d\!:\ \frac{1}2\,\equiv\, -k\,\equiv\, \frac{4}9\ \Rightarrow\ \frac{\color{#0a0}9}{18}\equiv \frac{\color{#c00}8}{18}\qquad\qquad$$

The fractions uniquely exist mod $\,d\,$ since $\,d\,$ is coprime to $\,2,3\,$ so $\,2,3\,$ are invertible mod $\,d\,$ (else $\,2\mid d\mid 2k+1\,\Rightarrow\,2\mid 1,\ $ or $\ 3\mid d\mid 9k+4\,\Rightarrow\,3\mid 4,\,$ contradiction).

Bill Dubuque
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There is no contradiction. Unless you believe that you can prove $\gcd(15,21)=1$ as well.

I suggest you rather note $d\mid(9k+4)-2\cdot (4k+1)=k+2$ and then $d\mid (4k+1)-4\cdot(k+2)=-7$. Also note that $k=5$ leads to $\gcd(49,21)=7$, i.e. the problem statement is wrong anyway.

Hagen von Eitzen
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  • Typo has been corrected $(4k+1\to 2k+1)$ so this answer no longer applies. Please delete it so as not to confuse readers. – Bill Dubuque Nov 24 '21 at 19:51
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Problem statement is wrong because for $k=7n+5,$ $n\geqslant 0$ we have that $\text{gcd}(9k+4,4k+1)$ can be equal to 7.

RFZ
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  • Typo has been corrected $(4k+1\to 2k+1)$ so this answer no longer applies. Please delete it so as not to confuse readers. – Bill Dubuque Nov 24 '21 at 19:51
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Note that if $d\ | 9k + 4$, and $d \ | 2k + 1$ then $d\ | (9k + 4) - 4 \cdot (2k + 1) = k$.

But

$\gcd(k, 2k + 1) = 1 \Rightarrow d = k = 1$

This implies $\gcd(9k + 4, 2k + 1) = 1$

Ps The fact that $\gcd(k, 2k+1) = 1$ is easy to see; any number $p$ that divides $k$, also divides $2k$ but do not divide $2k + 1$

Ant
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If $d$ is a common divisor of $a$ and $b$, then $d$ is a common divisor of any (integral) linear combination of $a$ and $b$. If we want to compute $\gcd(9k+4,2k+1)$, a first step might be to find a simple linear combination of $9k+4$ and $2k+1$. But we can take a linear combination that eliminats $k$ completely, namely

$$9(2k+1)-2(9k+4)=18k+9-18k-8=1.$$

Therefore any common divisor must divide $1$, and hence $\gcd(9k+4,2k+1)=1$

Aaron
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Please excuse me for this messed up question,
but I've come up with an answer as compensation:

Just apply the Euclidean algorithm, and in 3 lines you get 1 as GCD.

355durch113
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