Integer equalities persist as congruences $\!\bmod n$

Question

I am told that if $gcd(a,m)=1$ one then there exists integers $x,y$ such that $ax+my=1$ and therefore $ax≡ax+my≡1$ (mod m).

Could someone please explain to me why $ax+my=1$ implies $ax≡ax+my≡1$ (mod m)?

Answer 1

Equalities in $\,\Bbb Z\,$ always persist as congruences since

$$x = y\,\Rightarrow m\mid x-y\,\Rightarrow\, x\equiv y\!\!\!\pmod m$$

More generally if $\,m\mid n\,$ then congruences mod $\,n\,$ persist mod $\,m\,$ since

$$ x\equiv y\!\!\!\pmod n,\ \ m\mid n\mid x-y\,\Rightarrow\, m\mid x-y\,\Rightarrow\, x\equiv y\!\!\!\pmod m$$

OP is special case $\,n = 0\ $ (by $\,x\equiv y\pmod{\! 0}\!\iff\! 0\mid x-y\iff x=y)$.

Further, by the Polynomial Congruence Rule, roots $\!\bmod n\,$ of a polynomial (with integer coef's) also persist as roots $\!\bmod m\,$ when $\,m\mid n.\,$ In particular $\,(n=0)\,$ integers roots persits as roots $\!\bmod m\,$ for all $\,m$.