Find the least four-digit solution \(r\) of the congruence \(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55} \) .

tertre
Mar 23, 2017

#3**0 **

You have made a small mistake Max :)

\((r+2)^2=(r+1)^2\\ r+2=\pm (r+1)\\ \text{clearly r+2 can not equal r+1}\\ r+2=-(r+1)\\ r+2=-r-1\\ 2r=-3\\ r=-1.5\)

It would have been easier just to solve it as it was.

\(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55}\\ 2r \equiv-3 \pmod{55}\\ r \equiv -1.5 \pmod{55}\)

Melody
Mar 23, 2017