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Properties of
Congruences
If a =b mod m, a EZb mod n, and m and n are relatively prime, then
a -b mod mn. (See Property 5 of divisibility in 51.2.)
Proposition 1.3.1. The elements of Z/nsZ which have multiplicative
inverses are those which are relativelyprime to m, i.e., the numbers a for
which there exists b with ab z 1 mod m are precisely those a for
which g.c.d.(a, m) = 1. In addition, if g.c.d.(a,nt) = 1, then such an inverse
b can be found in 0(log3m) bit operations.
Proof. First, if d = g.c.d.(a,m) were greater than 1, we could not have
ab -1 mod m for any b, because that would irrlply that d divides ah - 1
and hence divides 1. Conversely, if g.c.d.(a,rn) = 1, then by Property 2
above we may suppose that a < m. Then, by Proposition 1.2.2, there exist
integers u and v that can be found in 0(log"7n) bit operations for which
ua +vm = 1. Choosing b = u, we see that m(1- UCL= 1 - ab, as desired.
Remark. If g.c.d.(a, m) = 1, then by rlcgabive powers a-n mod rn we
mean the n-th power of the inverse residue class, i.e., it is represented by
the n-th power of any integer b for which ah = 1 mod m.
Example 1. Find 160-' mod 841, i.e., the inverse of 160 modulo 841.
which g.c.d.(a, m) = 1. In addition, if g.c.d.(a,nt) = 1, then such an inverse
b can be found in 0(log3m) bit operations.
Proof. First, if d = g.c.d.(a,m) were greater than 1, we could not have
ab -1 mod m for any b, because that would irrlply that d divides ah - 1
and hence divides 1. Conversely, if g.c.d.(a,rn) = 1, then by Property 2
above we may suppose that a < m. Then, by Proposition 1.2.2, there exist
integers u and v that can be found in 0(log"7n) bit operations for which
ua +vm = 1. Choosing b = u, we see that m(1- UCL= 1 - ab, as desired.
Remark. If g.c.d.(a, m) = 1, then by rlcgabive powers a-n mod rn we
mean the n-th power of the inverse residue class, i.e., it is represented by
the n-th power of any integer b for which ah = 1 mod m.
Example 1. Find 160-' mod 841, i.e., the inverse of 160 modulo 841.
Solution. By Exercise 6(c) of the last section, the answer is 205.
Corollary 1. If p is a prime number, then every nonzero residue class
has a multiplicative inverse which can be found in U(log") bit operations.
If p is prime, then every nonzero residue class has a multiplicative
inverse which can be found in 𝑂(log! 𝑝) bit operations.
Suppose we want to solve a linear congruence 𝑎𝑥 ≡ 𝑏 𝑚𝑜𝑑 𝑚, where
without loss of generality, we may assume that 0 ≤ 𝑎, 𝑏 < 𝑚. First,
if gcd 𝑎, 𝑚 = 1, then there is a solution 𝑥" which can be found in
𝑂(log! 𝑚) bit operations, and all solutions are of the form 𝑥 = 𝑥" + 𝑚𝑛
for n an integer. Next, suppose that d = gcd 𝑎, 𝑚 . There exists a
solution if and only if 𝑑|𝑏, and in that case our congruence is
equivalent (in the sense of having the same solutions) to the
congruence 𝑎′𝑥 ≡ 𝑏′ 𝑚𝑜𝑑 𝑚′ : where 𝑎!
=
"
#
, 𝑏!
=
$
#
, 𝑚!
=
%
#
.
Now suppose that d = gcd 𝑎, 𝑚 . We need to prove that There
exists a solution if and only if 𝑑|𝑏, and in that case our
congruence is equivalent to the congruence 𝑎′𝑥 ≡ 𝑏′ 𝑚𝑜𝑑 𝑚′ :
where 𝑎!
=
"
#
, 𝑏!
=
$
#
, 𝑚!
=
%
#
.
Systems of congruence
Systems of congruence

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Systems of congruence

  • 2. If a =b mod m, a EZb mod n, and m and n are relatively prime, then a -b mod mn. (See Property 5 of divisibility in 51.2.) Proposition 1.3.1. The elements of Z/nsZ which have multiplicative inverses are those which are relativelyprime to m, i.e., the numbers a for which there exists b with ab z 1 mod m are precisely those a for which g.c.d.(a, m) = 1. In addition, if g.c.d.(a,nt) = 1, then such an inverse b can be found in 0(log3m) bit operations. Proof. First, if d = g.c.d.(a,m) were greater than 1, we could not have ab -1 mod m for any b, because that would irrlply that d divides ah - 1 and hence divides 1. Conversely, if g.c.d.(a,rn) = 1, then by Property 2 above we may suppose that a < m. Then, by Proposition 1.2.2, there exist integers u and v that can be found in 0(log"7n) bit operations for which ua +vm = 1. Choosing b = u, we see that m(1- UCL= 1 - ab, as desired. Remark. If g.c.d.(a, m) = 1, then by rlcgabive powers a-n mod rn we mean the n-th power of the inverse residue class, i.e., it is represented by the n-th power of any integer b for which ah = 1 mod m. Example 1. Find 160-' mod 841, i.e., the inverse of 160 modulo 841.
  • 3. which g.c.d.(a, m) = 1. In addition, if g.c.d.(a,nt) = 1, then such an inverse b can be found in 0(log3m) bit operations. Proof. First, if d = g.c.d.(a,m) were greater than 1, we could not have ab -1 mod m for any b, because that would irrlply that d divides ah - 1 and hence divides 1. Conversely, if g.c.d.(a,rn) = 1, then by Property 2 above we may suppose that a < m. Then, by Proposition 1.2.2, there exist integers u and v that can be found in 0(log"7n) bit operations for which ua +vm = 1. Choosing b = u, we see that m(1- UCL= 1 - ab, as desired. Remark. If g.c.d.(a, m) = 1, then by rlcgabive powers a-n mod rn we mean the n-th power of the inverse residue class, i.e., it is represented by the n-th power of any integer b for which ah = 1 mod m. Example 1. Find 160-' mod 841, i.e., the inverse of 160 modulo 841. Solution. By Exercise 6(c) of the last section, the answer is 205. Corollary 1. If p is a prime number, then every nonzero residue class has a multiplicative inverse which can be found in U(log") bit operations.
  • 4.
  • 5. If p is prime, then every nonzero residue class has a multiplicative inverse which can be found in 𝑂(log! 𝑝) bit operations.
  • 6. Suppose we want to solve a linear congruence 𝑎𝑥 ≡ 𝑏 𝑚𝑜𝑑 𝑚, where without loss of generality, we may assume that 0 ≤ 𝑎, 𝑏 < 𝑚. First, if gcd 𝑎, 𝑚 = 1, then there is a solution 𝑥" which can be found in 𝑂(log! 𝑚) bit operations, and all solutions are of the form 𝑥 = 𝑥" + 𝑚𝑛 for n an integer. Next, suppose that d = gcd 𝑎, 𝑚 . There exists a solution if and only if 𝑑|𝑏, and in that case our congruence is equivalent (in the sense of having the same solutions) to the congruence 𝑎′𝑥 ≡ 𝑏′ 𝑚𝑜𝑑 𝑚′ : where 𝑎! = " # , 𝑏! = $ # , 𝑚! = % # .
  • 7.
  • 8. Now suppose that d = gcd 𝑎, 𝑚 . We need to prove that There exists a solution if and only if 𝑑|𝑏, and in that case our congruence is equivalent to the congruence 𝑎′𝑥 ≡ 𝑏′ 𝑚𝑜𝑑 𝑚′ : where 𝑎! = " # , 𝑏! = $ # , 𝑚! = % # .