How do I prove exchangeable modularity? How do I prove that, considering all numbers natural, and p and i relatively prime,
$mp+n \not \equiv 0 \pmod i$
is the same as
$m-x \not \equiv 0 \pmod i$
considering x a natural number and the solution of
$xp+n \equiv 0 \pmod i$
?
 A: Since $p$ and $i$ are relatively prime, we know that $p$ has a multiplicative inverse mod $i$, which is to say there is a number $p^{-1}$ such that $p\,p^{-1}\equiv 1\pmod i$. If $x$ is a solution to
$$
xp+n\equiv 0\pmod i
$$
we can multiply both sides of the congruence above by $p^{-1}$ to obtain
$$
x+p^{-1}n\equiv 0 \pmod i
$$
Since we are also given that $m \not \equiv x\pmod i$ we have
$$
m+p^{-1}n \not \equiv 0 \pmod i
$$
Now, multiplying both sides of this by $p$ we obtain
$$
mp+n\not \equiv 0 \pmod i
$$
as required. This argument is reversible, so the two conditions in the original probem are equivalent.
A: $$xp+n\equiv 0\pmod i\,\wedge\,m-x\equiv 0\pmod i\,\Longrightarrow \,m\equiv x\pmod i\,\Longrightarrow$$$$\Longrightarrow\,mp+n\equiv xp+n\pmod i\equiv 0 \pmod i $$and you can go back and forth with the arrows above
A: Consider the two statements $mp+n \not\equiv 0 \pmod i$ and $xp+n \equiv 0 \pmod i$ and suppose that $m-x \equiv 0 \pmod i$
Since $m-x \equiv 0 \pmod i$ we know that $i\mid(m-x) \implies i\mid(m-x)k \ \ \ \  \forall k \in \mathbb{Z}$
Also, $(mp+n)-(xp+n)\not\equiv 0 \pmod i$ $\implies mp-xp \not\equiv 0 \pmod i$
Therefore $(m-x)p \not\equiv 0\pmod i \implies i \not\mid (m-x)p$
This is a contradiction, therefore  $mp+n \not\equiv 0 \pmod i$ and $xp+n \equiv 0 \pmod i$ $\implies$ $m-x \not\equiv 0 \pmod i$
A: Hint $\rm\,\ 0\not\equiv mp\!+\!n \equiv mp\!+\!n\!-\!(xp\!+\!n)\equiv (m\!-\!x)\,p.\:$ Scale this by $\rm1/p\:$ (exists by $\rm\:(p,i)=1).$
