gcd(2a+1, 9a+4) = 1

Question

The question is from Burton's Elementary Number Theory. I want to know if my proof is legible. Proof : Let $$d=gcd(2a+1, 9a+4)$$ Then $$d|2a+1$$ and $$d|9a+4$$ $$2a+1=db$$ and $$9a+4=dc$$ $$ a = \frac{db-1}{2}$$and$$a= \frac{dc-4}{9}$$ Equating both equations : $$ 9db-9=2dc-8 $$ $$ d(9b-2c) = 1 $$ $$ 9b-2c = \frac{1}{d}$$ Now, since b and c are integers, therefore $\frac{1}{d}$ is an integer, i.e d divides 1 and therefore $$gcd(a, b)= d = 1$$.

It's correct. Essentially you eliminated $,a,,$ which can be done more directly as follows:
$$\bmod d!:,\ 2a!+!1\equiv 0\equiv 9a!+!4\ \Rightarrow\ \color{#c00}0\equiv 9(2a!+!1)-2(9a!+!4)\equiv\color{#c00} 1\ \Rightarrow\ d\mid \color{#c00}{1!-!0}\qquad$$ — Bill Dubuque, Jan 04 '19 at 17:02
@BillDubuque Thanks. I'm sure it's a more direct approach but I'm unfamiliar with modular arithmetic so it had to be something without it. — P.Jo, Jan 04 '19 at 17:22
We can eliminate congruence notation too! Namely the above means
$$d\mid 2a!+!1,9a!+!4\ \Rightarrow\ d\mid 1!=! 9(2a!+!1)-2(9a!+!4)\qquad $$ since multiples of $d$ are closed under integral linear combinations. — Bill Dubuque, Jan 04 '19 at 17:27
@BillDubuque Yeah. That makes perfect sense. But what does it mean to be 'closed under integral linear combinations'? — P.Jo, Jan 05 '19 at 04:19
@BillDubuque Does it mean that the multiples of d= gcd(a, b) are all the linear combinations of a and b? — P.Jo, Jan 05 '19 at 04:27
It means $,d\mid a,b,\Rightarrow, d\mid ja+kb,$ for all integers $,j,k.\ $ It is true that $, \gcd(a,b)\Bbb Z = a\Bbb Z + b\Bbb Z, $ by Bezout. — Bill Dubuque, Jan 05 '19 at 04:43
@BillDubuque And what does $, \gcd(a,b)\Bbb Z = a\Bbb Z + b\Bbb Z, $ mean? — P.Jo, Jan 05 '19 at 12:49
It means $ { an + bm\ :\ n,m\in\Bbb Z} = { dn\ :\ n\in\Bbb Z},,$ i.e. the set of integral linear combinations of $,a,,b,$ equals the set of all multiples of their gcd $d.,$ More generally any nonzero set of integers closed under subtraction equals the set of multiples of its least positive element $d$ (which is the gcd of all its elements). This is clarified whe one studies (cyclic) groups and (principal) ideals and Euclidean domains. — Bill Dubuque, Jan 05 '19 at 14:32

Wuestenfux · Answer 1 · 2019-01-08T08:36:52.160

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Divide $2a+1$ into $9a+4$ giving $9a+4 = 9/2\cdot (2a+1) - 1/2$ or $2(9a+4) = 9(2a+1)-1$, or $1= -2(9a+4) + 9(2a+1)$. So each divisor of $9a+4$ and $2a+1$ divides $1$.

edited Jan 08 '19 at 08:36

answered Jan 04 '19 at 15:39

Wuestenfux

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You are dividing $2a+1$ into $9a+4$, not the other way around as you have typed. – Michael Munta Jan 07 '19 at 18:27
Indeed, you're right. Thanks. – Wuestenfux Jan 08 '19 at 08:37

score 1 · Answer 2 · answered Jan 04 '19 at 15:39

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Looks good to me. Some nit picks follow:

A bit of formatting may help readability. For much of your answer you have pairs of equations which belong together, but the pairs which belong together aren't the pairs which appear close to one another.

After "therefore $\frac{1}{d}$ is an integer", you can just say "so $d=1$ and we're done". Remember that $d=1$ is the very thing we want to prove, so it's unnecessary to continue with more calculations once you've reached that conclusion.

And personally I would probably have used the Euclidean algorithm instead.

answered Jan 04 '19 at 15:39

Arthur

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The formatting is sloppy indeed. Thanks for the 'nitpicks'. I seem to lack the knowledge of the Euclidean Algorithm since it is the next topic in the book. – P.Jo Jan 04 '19 at 16:21
@P.Jo Then it's no wonder you didn't use it. The Euclidean algorithm is basically a way of simplifying the two expressions without changing their $\gcd$. If you can simplify until one of them becomes $1$ or $0$ (which always happens if you only have integers, but not necessarily if you have expressions like yours), then you're done. If not, then at least it will be simplified and easier to use alternate methods like yours. – Arthur Jan 04 '19 at 16:30
Nice preview before starting. Sounds like it's gonna make this easier. – P.Jo Jan 04 '19 at 17:11

score 1 · Answer 3 · answered Jan 11 '19 at 19:25

As Siong Thye Goh said, we can use the Euclidean algorithm to guide us to a proof. The Euclidean algorithm is based on the fact that $gcd(x,y)=gcd(x,y-tx)$. If we take $x=2a+1$, $y=9a+4$, and $t=4$, then we get $gcd(2a+1,9a+4)=gcd(2a+1,a)$. Applying it again with $x=a$, $y=2a+1$, $t=2$, we get $gcd(2a+1,a)=gcd(1,a)$. From there, it's easy to see that the gcd is $1$.

score 0 · Accepted Answer · answered Jan 04 '19 at 15:42

0

Seems fine.

Alternatively,

$$9a+4=4(2a+1)+a$$

$$2a+1=2(a)+1$$

Hence,

$$1=(2a+1)-2a=(2a+1)-2(9a+4)+8=9(2a+1)-2(9a+4)$$

That is the greatest common divisor of $2a+1$ and $9a+4$ must divide $1$. Hence the greatest common divisor is $1$.

answered Jan 04 '19 at 15:42

Siong Thye Goh

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What did you do here really? – Michael Munta Jan 06 '19 at 10:57
I am using Euclidean algorithm to express the gcd as a combination of the two numbers. – Siong Thye Goh Jan 06 '19 at 11:08
Can you please explain how this happens? I know how it works with integers, but not with expressions like these. – Michael Munta Jan 06 '19 at 11:16
For the first part, I divide $9a+4$ by $2a+1$ and find that the remainder to be $a$. After which , I divide $2a+1$ by $a$ and obtain a remainder of $1$, then I work backward to substitute $a$ in terms of $9a+4$ and $2a+1$. – Siong Thye Goh Jan 06 '19 at 11:21
This is the extended algorithm? – Michael Munta Jan 06 '19 at 12:36
Ah yes. I should have included the word "extended". – Siong Thye Goh Jan 06 '19 at 13:11
How did you get $4(2a+1) + a$ by dividing $9a + 4$ by $2a+1$? Can you show the steps please? – Michael Munta Jan 06 '19 at 14:04
Let us continue this discussion in chat. – Siong Thye Goh Jan 06 '19 at 14:26
Was that polynomial long division? – Michael Munta Jan 06 '19 at 22:04

gcd(2a+1, 9a+4) = 1

4 Answers4