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I am trying problems from Apostol Modular functions and Dirichlet series in number theory and I could not think about this problem from chapter 2 .

Problem is – Given integers $a, b, c, d\;$ with $ad-bc \equiv 1 \pmod n$, prove that there always exists integers $\alpha,\beta, \gamma, \delta$ such that $\alpha \equiv a \pmod n$, $\beta \equiv b \pmod n$, $\gamma \equiv c \pmod n$, $\delta \equiv d \pmod n$ with $\alpha \delta-\beta \gamma = 1$ .

I am unable to think how to prove existence of $\alpha, \beta, \gamma, \delta$ which are equivalent to $a, b, c, d \bmod n$ respectively. Can someone please give hints.

Bill Dubuque
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  • Please read again my comment on your previous question: https://math.stackexchange.com/questions/3483436/doubt-in-a-proof-of-an-exercise-in-apostol-modular-functions-and-dirichlet-serie – Angina Seng Dec 21 '19 at 14:05
  • @Lord Shark the Unknown I know mathjax, the problem is my browser doesn't have preview post option, so I can't see what could be possible mistake. –  Dec 21 '19 at 14:07
  • @Lord Shark the Unknown so, I have to edit posts after posting question. I have edited it. –  Dec 21 '19 at 14:08
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    You are still putting individual symbols between dollar signs rather than the whole formula; not only does that look awful, it's more work than doing it properly. – Angina Seng Dec 21 '19 at 14:10

1 Answers1

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$(\overbrace{a\!+\!\ell n}^{\textstyle \alpha},\:\!b)=1\,$ for $\,\ell\in\Bbb Z\,$ by $\,{(a,b,n)=1,}\,$ by here. Let $\,\beta =b.\,$ We solve for $\,\delta,\gamma\in\Bbb Z$.

$\!\!\bmod n\!:\ \color{#0a0}{\alpha\equiv a}\,\Rightarrow \color{#0a0}\alpha d\!-\!b c = \color{#0a0}ad\!-\!bc = 1,\ $ so $\,\alpha d\! -\! b c = \color{#c00}{1\!-\!kn}\,$ for $\,k\in\Bbb Z,\,$ hence

$1\! =\! \underbrace{\alpha(d\!+\!in)\!-\!b(c\!+\!jn)}_{\textstyle \alpha\,\delta \,-\, \beta\,\gamma}\! =\! \color{#c00}{1\!-\!kn}\!+\!(i\alpha\!-\!jb)n\!\iff\! i\alpha\!-\!jb = k.\,$ Such $\,i,j\,$ exist by $(\alpha,b)\!=\!1$

Bill Dubuque
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  • where did you used the hyperlink 'here'. Also, can you please tell as (a, b) =1 but why integers have the form d+in, c+jn in 1= a(d+in) -b ( c+jn) ? –  Dec 22 '19 at 06:27
  • can you please elaborate your answer, I am sorry but i am confused. –  Dec 22 '19 at 06:31
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    @user686624 I expanded it. Note $,\delta \equiv d\pmod{!n},$ $\Rightarrow,\delta = d + in,$ for $, i\in \Bbb Z,,$ and similarly $,\gamma\equiv c\pmod{!n},$ $\Rightarrow, \gamma = c + jn,$ for $,j\in\Bbb Z.,$ Above we show that such $,i,j,$ exist satisfying $,\alpha \delta - \beta \gamma = 1\ \ $ – Bill Dubuque Dec 22 '19 at 08:30
  • which result are you using to prove existence of such i, j? gcd of $\alpha$ , b is 1 but k not necessarily equal to 1? –  Dec 22 '19 at 11:27
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    @user686624 The well-known basic result that $,\exists, x,y\in\Bbb Z!:\ ax+by=c\iff (a,b)\mid c\ $ or, equivalently, $,\exists, x!:\ ax\equiv c\pmod{!b}\iff (a,b)\mid c\ \ \ $ – Bill Dubuque Dec 22 '19 at 19:04
  • got it. Can you please look at the problem ( that I have mentioned in this question by editing) . It is related to the problem whose solution ( you have wrote. –  Dec 22 '19 at 19:23
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    @user686624 If you have another question you need to post it separately (and link to this if it is related). – Bill Dubuque Dec 22 '19 at 19:26
  • Ok I will do that –  Dec 22 '19 at 19:29