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Allow $a,b,c \in \mathbb{Z}$ and $a,c<b$. Does there exist a $b$ such that, for fixed $c$ and $1\leq a \leq b-2$, $a*c \mod b \equiv a+1 \mod b$? For example, $2*c \mod b \equiv 3 \mod b$, $3*c \mod b \equiv 4 \mod b$, and so on up to $(b-2)*c \mod b \equiv b-1 \mod b$.

Some things to note:

  • If we allow $1\leq a \leq b-1$, then the only numbers satisfying this property are $b=c=2$. Because of the way we defined our domain for $a$, we imply that $(b-1)*c \mod b \equiv b \mod b \equiv 0 \mod b$. This is only possible if $b-1 | b$, which implies $b=2$. Our only possible value of $a$ satisfying $1\leq a \leq 1$ is $a=1$, and sure enough, $1*2 \mod 2 \equiv 2 \mod 2 \equiv 0 \mod 2$.
  • Next, we observe the fact that if $b$ exists, it must be odd, since $2*c \mod b \equiv 3 \mod b$. If $b$ were even (and $b=2k$), then this would imply $2c+2mk = 3 + 2nk$ for some $m,n\in \mathbb{Z}$. This clearly can never be the case.
  • Now, note that $c$ would have to be relatively prime to $b$. Otherwise, then if $\frac{b}{c} = d$, clearly $d\in \mathbb{Z}_b$, and $c*d \mod b \equiv 0 \mod b$ instead of $d+1 \mod b$.
Bill Dubuque
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AKemats
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  • It's trivial: $,\bmod b!:,\ \color{#c00}{c\equiv 2 },\ 2\color{#c00}c \equiv 3\Rightarrow 2(\color{#c00}2)\equiv 3\Rightarrow 1\equiv 0\Rightarrow b\mid 1\ \ $ – Bill Dubuque Jan 19 '22 at 18:37
  • Please, avoid using $*$ to denote multiplication, this is Mathematics forum, not a programming forum. – jjagmath Jan 19 '22 at 19:57
  • Per my prior comment, this is an immediate consequence of basic congruence arithmetic laws so I am going to close this as a dupe of these laws. I highly recommend you learn how to work with congruence relations rather than mod operators - e.g. see here. Using mod operations may have caused this to not seem as trivial as it is. – Bill Dubuque Jan 20 '22 at 09:24

1 Answers1

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No, there are no numbers which satisfy this property.

The final nail in the coffin is this: since $1*c \mod b \equiv 2 \mod b$, $c=2$. What we want is $2*c \mod b \equiv 3 \mod b$, but what we have is $4 \mod b \equiv 3 \mod b$. The only way this is possible is if $b = 1$, and there are no numbers which satisfy the inequality $1 \leq a \leq 1-2=-1$.

AKemats
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