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# question

0
287
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+3031

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
#1
+7023
+1

$$(r+2)^2\equiv (r+1)^2 \pmod{55}\\ r+2\equiv r+1\pmod{55}$$

what?????

MaxWong  Mar 23, 2017
#2
+3031
+1

someone help!

tertre  Mar 23, 2017
#3
+93307
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
#4
+7023
0

The solution of r should be a 4-digit solution, as the question requires......

MaxWong  Mar 23, 2017
#5
0

The smallest 4-digit solution is when:

r=81, so the LHS of the equation is:

6,889 mod 55 = 14, and the RHS of the equation is:

6,724 mod 55 = 14.

Guest Mar 23, 2017