$a \equiv b \mod k \implies a \equiv ? \mod k^2$?

Question

Is there any modular law such that:

$a \equiv b \mod k \implies a \equiv ? \mod k^2$

I know that $$ a \equiv b \mod k \implies a^n \equiv b^n \mod k $$ but sometimes I need to power only $k$ and then there are examples when it is not so easy to investigate without any theorem.

Answer 1

The left hand side means $k\mid(a-b)$. Square to get $k^2\mid (a-b)^2$. You can write that as $a\equiv a^2+(1-2b)a+b^2\bmod{k^2}$, though it is a bit artificial.

Answer 2

Since you only know $a\equiv b\mod k$, then $a=b+jk$ for some $j$. Now, this last expression mod $k^2$ depends only in $j\mod k$, what I am trying to say is $b+jk\equiv b+j'k\mod k^2$ if $j\equiv j'\mod k$. So, what we can say is that $a\equiv b+jk\mod k^2$ for some $j\mod k$, in better words: $$a\equiv b+jk\mod k^2\hbox{ for some } j=0,1,\dots ,k-1$$