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Suppose

$$ac\equiv -b \bmod{n}.$$ Let $c^{-1}$ be modular multiplicative inverse of $c$ modulo $n$.

Eventually, Why we can conclude that

$$a\equiv -bc^{-1} \bmod{n} \;?$$

I read this page but I couldn't found a helpful description about my issue.

Gary
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tstt
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  • By the Congruence Product Rule in the linked dupe, congruences are preserve by scalings, so scaling your first congruence by $c^{-1}$ yields your second congruence. Note $c$ is invertible $!\bmod n\iff c$ is coprime to $n\ \ $ – Bill Dubuque Dec 31 '22 at 07:54
  • Specifically, apply the following instance of the Product Rule: $$\bmod n!:,\ A\equiv B\ \ &\ \ C\equiv C\ \Rightarrow\ {AC\equiv BC}\qquad$$ where $,A = ac,\ B = -b,\ C = c^{-1}.\ $ It's the congruence analog of the fact that equation are preserved under scalings (or application of any function) – Bill Dubuque Dec 31 '22 at 08:01
  • @BillDubuque $AC=ac\frac{1}{c}$? – tstt Dec 31 '22 at 08:05
  • $AC = (ac)c^{-1} \equiv a(cc^{-1})\ \equiv a(1)\equiv a.\ $ The idea is we are dividing both sides by $c$ by scaling by $c^{-1},,$ just like for normal equations (congruences are generalized equations). – Bill Dubuque Dec 31 '22 at 08:07
  • You can use fractions in modular arithmetic such as $c^{-1} \equiv \dfrac{1}{c}$ but you need to be extra careful to ensure you don't divide by zero (or a zero-divisors), i.e. we can only divide by invertibles, so denominators must be coprime to the modulus. – Bill Dubuque Dec 31 '22 at 08:11
  • @BillDubuque I read this from MIT slides: we have $$I); a_0(x_0-y_0)\equiv -\sum_{i=1}^ra_i(x_i-y_i) \mod m$$ since $$II);;(x_0\neq y_0)$$ an inverse $(x_0-y_0)^{-1}$ must exist, which implies that $$ III);;a_0\equiv -\sum_{i=1}^ra_i(x_i-y_i)(x_0-y_0)^{-1} \mod m$$ My problem is how we can achieve by this assumption that $I,II$ be true? – tstt Dec 31 '22 at 08:24
  • $c := x_0-y_0$ is invertible $!\bmod m\iff c,$ is coprime to $m\ \ $ – Bill Dubuque Dec 31 '22 at 09:00

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