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2 0 What I have done is this: $aequiv b pmod{2n}$, $a=b+ctimes2n$, for some $c$, $a^2=b^2+2btimes ctimes2n+c^2times2^2n^2$, $a^2-b^2=(btimes c+c^2n)times4n$, then $a^2equiv b^2pmod{2^2n}$. I think that this is right: what I DON’T understand is how to generalize this to: $aequiv bpmod{kn}Rightarrow a^kequiv b^k pmod{k^2n}$. Please give me a hint. number-theory share | cite | improve this question edited Jan 4 at 15:33 Bill Dubuque 209k 29 190 631 asked Dec 1 '11 at 10:52 ...