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I am self studying number theory from Introduction to Analytic number theory by Apostol.

I have a doubt in one argument of proof of Theorem 5.14 and I am posting its image.

My doubt is in 2nd line of 2 nd paragraph how Apostol writes - If y is a solution of (5) then ay $\equiv $ at ( mod m) . How does y being a solution of (5) implies that? enter image description here

Can someone please help

Bill Dubuque
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    $ay\equiv b\equiv at\pmod m$ because $y,t$ are, by assumption, solutions of (5). – Fabio Lucchini Dec 26 '19 at 19:15
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    I added an answer that may help to clarify the idea behind the proof. I can elaborate on that if need be - just post questions in comments. – Bill Dubuque Dec 26 '19 at 20:55
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    I'm voting to close this question as off-topic because the person who asked it has abandoned it. – José Carlos Santos Dec 29 '19 at 09:35
  • I don't know why $ax\equiv at (\mod m)\Rightarrow x\equiv t(\mod m/d).$ A book tells me that, $m| a(x - t)\Rightarrow m | d(x - t)$ because if $b | ac,$ then $b | (a, b)c.$ This is because if $b | mn$ and $(b, m) = 1,$ then $b | n.$ This is because $1 = bp + mq\Rightarrow n = bnp + mnq.$ As $b | (bnp + mnq)\Rightarrow b | n.$ –  Nov 27 '21 at 11:46
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    @LuânTăngVăn It has been answered before and I have accepted the answer. –  Nov 27 '21 at 13:08

1 Answers1

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$t\,$ is a root of $(7),$ which, scaled by $d,\,$ shows $t$ is root of $(5)$, so $\,at\equiv b\equiv ay\Rightarrow at\equiv ay,\,$ by transitivity of congruence (being an equivalence relation).

The proof is poorly presented. Below is a more conceptual presentation that generalizes widely. Like linear differential and difference equations (recurrences), for any linear congruence $\,ax \equiv b,\,$ it's easy to show (see here or here) the $\rm\color{darkorange}{general}$ solution is the sum of any $\color{#0a0}{{\rm particular\ solution}\ x_0}\,$ plus the general solution of associated homogeneous equation $ax \equiv 0\pmod{\!m},\,$ which here is $\, m\mid ax\!\iff m/d\mid (a/d)x\!\iff m/d\mid x,\,$ by Euclid & $\,(m/d,a/d)\! =\! (m,a)/d = 1.\,$ Viewed $\!\bmod m\ $ such $\:\!x\:\!$ are the $\,\color{#c00}d\,$ multiples of $\,m/d,\,$ i.e. $\, m/d\,\color{#c00}{\{0,1,2,\ldots, d\!-\!1\}}^{\phantom{|^|}}\!$ [to be rigorous note $\!\bmod m\!:\,\ x\,\equiv\, k(m/d)\,\equiv\, k(m/d)\bmod m\:\overset{\small\rm\color{#90f}{DL}}\equiv\: m/d\,\color{#c00}{(k\bmod d)}$ by ${\small\rm\color{#90f}{DL}}\!=\!$ $\!\bmod\!$ Distrib. Law].
Hence the $\rm\color{darkorange}{general}$ solution is $\!\bmod m\!:\, x\equiv \!\!\!\!\!\!\!\!\underbrace{\color{#0a0}{x_0} + m/d\ \color{#c00}k^{\phantom{|^{|^.}}\!\!\!}}_{\text{$\rm \color{#0a0}{particular}$ + homogeneous}}\!\!\!\!\!\!\!\!\!,\ \color{#c00}{0\!\le\! k\!<\! d}\ \ $ [$x_0 = t\,$ in the OP]

Remark $ $ Said structure of the solution space will be clarified when one studies linear algebra and modules (= linear algebra where the coefficient algebra is a ring vs. field)

See here for more on switching back-and-forth between the equivalent linear congruence and its associated Bezout linear equation, and links to handy algorithms for solving such equations.

See here for a convenient view of the above in terms of modular (multi-valued) fractions

Bill Dubuque
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