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I was looking for proof of the two problems like this.

The first one is if $$\gcd(a, b) = 1$$ then $$\gcd\left(a, (b^{-1}\ \rm{mod}\ a)\right)=1$$ which $(b^{-1}\ \rm{mod}\ a)$ is the modular multiplicative inverse of $b$ in $\rm{mod}\ a$ .

And the second one is if $$\gcd(a, b_n) = 1\ \text{ and }\ b_n = sb_i\ \text{ and }\ B = b_n\times(b_n^{-1}\ \rm{mod}\ a)$$ then $$\gcd(b_i, \frac{B-1}{a})=1$$

In order to proof this problems I did like this $$A=(b^{-1} \mod\, a)$$ $$bk \equiv 1 \mod a$$ $$k \equiv A \mod a$$ so $$k=ak_1+A $$ now if $A=ak_2$

then $$k=ak_1+ak_2=a(k_1+k_2)=ak_3$$ but $$k=ak_1+A$$ as $$A\neq 0$$ so it's a contradiction as a result $$A\neq ak_1$$ $$\gcd\left(a, A\right)=1$$ $$\gcd\left(a, (b^{-1}\ \rm{mod}\ a)\right)=1$$

for the second one I did like this: $$B=b_nA = b_i\times sA = b_i k$$ so $$\frac{B-1}{a}=\frac{b_i k-1}{a}=D$$ now if $D=b_ik_0 \text{ for } k_0 \in Z$ $$\frac{b_i k-1}{a}=b_ik_0$$ $$\frac{b_i k-1}{ab_i}=k_0=\frac{k-1}{ab_i}-\frac{1}{ab_i}$$ so $$k_0 \notin Z$$ but we said that $k_0 \in Z $ and this one is contradiction too so $$D\neq b_ik_0$$ so $$gcd(D,b_i)=1$$

but I'm not sure if it is true or not!!

a.Tesla
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  • Please tell us how do you think about these problems and include your efforts in the post. – on1921379 Aug 13 '22 at 10:44
  • thanks for review , I edited the post and it's now include my proofs.@on1921379 – a.Tesla Aug 13 '22 at 12:14
  • thanks I fixed it .@PeterPhipps – a.Tesla Aug 13 '22 at 12:15
  • These are immediate consequences of the basic divisibility properties that are proved in the linked dupes. Let $,b' = b^{-1} \bmod a,,$ so $,bb'\equiv 1\pmod{a}\Rightarrow \color{#c00}b\color{#0a0}{b'}+\color{#0a0}a\color{#c00}n = 1.,$ Thus $,\color{#0a0}{(a,b')}=1,,$ and $, 1 =\color{#c00}{(b,n)} = (b,(bb'!-!1)/a) = (b,B!-!1)/a),,$ thus $,b_i\mid b \Rightarrow (b_i,B!-!1)/a)=1$ – Bill Dubuque Aug 13 '22 at 17:42

1 Answers1

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The proof of the first problem is as follows. $\gcd(a,b)=1$, so $$\gcd(a,(b^{-1}\mod a))=\gcd(a,b\cdot(b^{-1}\mod a))=\gcd(a,(1\mod a))=1.$$

Briar Jam
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