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Suppose $a$ is odd. I want to show that there are infinitely many $b$ for which $(a,b)=(a,b+1)=1$.

What I have done is this:

Let $b=ak+1$. Then $1=(-k)a+b$, and so $\gcd(a,b)=1$. If $2=(-k)a+b'$, then $\gcd(b',a)=1$.

First of all, is this correct?

Secondly, my book says to use Chinese Remainder Theorem. How to apply CRT to come up with $ak+1$ and $ak+2$?

Bill Dubuque
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violet
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1 Answers1

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Your proof is correct, but can be polished by saying after $2=-ka+(b+1)$ that since an integer linear combination of $a$ and $b+1$ sums to $2$, $(a,b+1)\mid2$, so that GCD is $1$ or $2$.

The book's proof begins with the prime power factorisation of $a$, noting that the factors will be at least $3$. By CRT we can compute infinitely many numbers $b$ that are $1$ modulo all of those factors, so $(a,b)=1$. But then $b+1$ is $2$ modulo all of those factors, so $(a,b+1)=1$ as well. This is a snazzier presentation than yours, but essentially the same content since $ak+1$ is $1$ modulo any factor of $a$ – your proof is more understandable.

Parcly Taxel
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  • Thanks. But is my proof correct? – violet Feb 22 '23 at 07:36
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    @violet It is correct, actually. – Parcly Taxel Feb 22 '23 at 07:43
  • @violet CRT is major overkill (not "snazzier") for this simple problem, since it can be simpler deduced as you did, i.e. $\color{#c00}{1,2}$ are coprime to odd $,a,,$ so $\bmod a!:\ b\equiv 1,\Rightarrow, b,b!+!1\equiv \color{#c00}{1,2},$ are also coprime to $,a\ $ (clearer by gcd mod reduction vs. your use of Bezout). – Bill Dubuque Feb 22 '23 at 20:56
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    @violet Said more simply: if $,b,b!+!1,$ are coprime to $,a,$ then so too are $,a!+!b,,a!+!b!+!1,,$ so continually adding $,a,$ yields infinitely many coprime pairs. – Bill Dubuque Feb 22 '23 at 21:37