+489 Find the least positive four-digit solution r of the congruence \(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55} \).
Are you sure your congruence is written correctly? With a modulus of 55, there doesn't appear to be a solution !!. It can be written as follows: (r + 2)^2 ≡(r + 1)^2 mod 55.
+118609 Find the least positive four-digit solution r of the congruence .
\(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55}\)
It can be written as follows: (r + 2)^2 ≡(r + 1)^2 mod 55. ( just as guest said )
(r + 2)^2 ≡(r + 1)^2 mod 55.
Let x=r+1
\((x+1)^2\equiv x^2 \pmod{55}\\ x^2+2x+1\equiv x^2 \pmod{55}\\ 2x+1\equiv 0 \pmod{55}\\ 2x+1=55k \qquad k\in Z\\ 2x=55k-1 \qquad k\in Z\\ \text{(x-1) must have 4 digits}\\ \text{ k=37 tis the smallest, k needs to be odd}\\ Try\;\; k=37, 39,41,43,.....\\ 2x+1= 2035, 2145, 2255, 2365,\dots\\ 2x=2034, 2144, 2254, 2364, \dots \\ x=1017,1072, 1127, 1182 \dots\\ r=1016, 1071, 1126, 1181 \dots (1016+55c) \qquad c\in Z \)
check
mod(1017^2,55) = 14
mod(1018^2,55) = 14
mod(1072^2,55) = 14
mod(1073^2,55) = 14
Well the first 2 work anyway , so it looks ok
+118609 I've done some homework this morning and I do not think it is quite right.
According to what I read this is only one answer. The same answer written in different ways.
Plus
I probably didn't need to make it look so complicated ...
Here is the answer that I found before, only I have done it much more simply.
\(r^2 + 4r + 4 \equiv r^2 + 2r + 1 \pmod{55}\\ 4r + 4 \equiv 2r + 1 \pmod{55}\\ 2r \equiv -3 \pmod{55}\\ 2r \equiv 52 \pmod{55}\\ r \equiv 26 \pmod{55}\\ r \equiv 26+55k \pmod{55}\\ r \equiv 26+1045 \pmod{55}\\ r \equiv 1071 \pmod{55}\\ \)
Now i need to find 3 other answers. (Assuming that if I just add multiples of 55 I will just keep getting the same answer expressed in different way.)
Mmm.....
I don't know about any other ones .... 
