Example:

Goldwasser-Micali with n=217 and y=145. 
Note 217 = 7 * 31. y=145 is a pseudosquare since 

  (y|7) = (145)^(3) = -1 (mod 7) and (y|31) = (145)^(15) = -1 (mod 31)

by using Euler's criterion: the Legendre symbol (y|p) is given by (y|p) = y^{(p-1)/2} (mod p).

Alice sent a 4-bit message encoded using the following numbers:

  x[1] = 185 (mod 217)
  x[2] = 144 (mod 217)
  x[3] = 61 (mod 217)
  x[4] = 211 (mod 217)

Bob decodes the bits by checking the Legendre symbol (x|p):

  (185 | 7) = 185^3 (mod 7) = -1 => m[1] = 1 that is 185 is a pseudosquare
  (144 | 7) = 144^3 (mod 7) = 1  => m[2] = 0 that is 144 is a square (quadratic residue)
  (61 | 7) = 61^3 (mod 7) = -1   => m[3] = 1 that is 61 is a pseudosquare
  (211 | 7) = 211^3 (mod 7) = 1  => m[4] = 0 that is 211 is a square (quadratic residue)

Note: (185 | 7) = (185 | 31) = -1, that is, it suffices for Bob to check one value of
the Legendre symbol (as they are supposed to agree).


Example: 

Blum-Goldwasser with n=217
Alice sent the ciphertext (1,1,1,1,1,1,1,1) and s[9]=39(mod 217).
What is her plaintext to Bob?

  n = p*q, where p = 7 and q = 31

Recall that the square root function for a prime congruent to 3 mod 4 is:
  sqrt(x) mod p = x^[(p+1)/4] (mod p)

Bob needs to compute s[1] mod 217 (by taking 8 consecutive square roots):
  s[1] = s[9]^u (mod 7)
  s[1] = s[9]^v (mod 31)

where
  u = [(7+1)/4]^8 (mod 6)   = 4
  v = [(31+1)/4]^8 (mod 30) = 16

Thus,
  s[1] = 4 (mod 7)
  s[1] = 8 (mod 31)

Apply the Aryabhata (extended Euclidean) algorithm on 7 and 31:
-------------------------------------------------------------
  A	B	Q	R	X1	Y1	X2	Y2
-------------------------------------------------------------
  31	7	4	3	1	0	0	1
  7	3	2	1	0	1	1	-4
  3	1	3	0	1	-4	-2	9
-------------------------------------------------------------
check: 1 = (-2)*31 + 9*7

By the Chinese Remainder Theorem:
  s[1] = 4*(5*31) + 8*(9*7) = 39 (mod 217)

Checking the Blum-Blum-Shub pseudorandom sequence and deciphering
the sequence (1,1,1,1,1,1,1,1):
  s[1] = 39	b[1] = 1	m[1] = 0
  s[2] = 2	b[2] = 0	m[2] = 1
  s[3] = 4	b[3] = 0	m[3] = 1
  s[4] = 16	b[4] = 0	m[4] = 1
  s[5] = 39	b[5] = 1	m[5] = 0
  s[6] = 2	b[6] = 0	m[6] = 1
  s[7] = 4	b[7] = 0	m[7] = 1
  s[8] = 16	b[8] = 0	m[8] = 1
  s[9] = 39 

Plaintext = 01110111