MHB Coprime mod n implies coprime-ish mod n

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The discussion centers on the relationship between integers \(a\) and \(b\) under modular arithmetic, specifically when \(ap + bq \equiv 1 \mod n\). It is established that if \(\gcd(a, b) = 1\), then integers \(a', b', p',\) and \(q'\) can be found such that \(a'p' + b'q' = 1\). The method involves adjusting \(p\) and \(q\) by adding multiples of \(n\) to maintain the modular equivalence. However, the implications when \(\gcd(a, b) \neq 1\) remain unclear. Overall, the discussion emphasizes the conditions under which coprimality influences modular relationships.
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Let $a$ and $b$ be two integers such that there exists integers $p$, $q$ with $ap+bq=1\text{ mod }n$. Do there exist integers $a^{\prime}$ $b^{\prime}$, $p^{\prime}$ and $q^{\prime}$ such that, $x^{\prime}=x\text{ mod }n$ for $x\in\{a, b, p, q\}$ and, $$a^{\prime}p^{\prime}+b^{\prime}q^{\prime}=1?$$
 
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Re: Coprime mod $n$ implies coprime-ish mod $n$

Swlabr said:
Let $a$ and $b$ be two integers such that there exists integers $p$, $q$ with $ap+bq=1\text{ mod }n$. Do there exist integers $a^{\prime}$ $b^{\prime}$, $p^{\prime}$ and $q^{\prime}$ such that, $x^{\prime}=x\text{ mod }n$ for $x\in\{a, b, p, q\}$ and, $$a^{\prime}p^{\prime}+b^{\prime}q^{\prime}=1?$$

If $\gcd (a,b)=1$ then yes.

$ap+bq \equiv 1 \mod n$ means there exist integer $\gamma$ such that $ap + bq+n \gamma =1$ . If $\gcd (a, b)=1$ then $\exists k_1, k_2$ such that

$ak_1+bk_2=\gamma$.

Take $a{'} =a, b{'}=b, p{'}=p+nk_1, q{'}=q+nk_2$

Then $a{'}b{'} +b{'}q{'}=1$

I am not sure what happens when $\gcd (a,b) \neq 1$.

Hope this helps.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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