Number Theory
Chapter 4 – Sections 2 and 3
Congruences
Congruences
Def: A congruence of the form
ax b (mod m),
where x is an unknown integer, is called a linear
congruence in one variable.
Note: If x0 is a solution, then ax0 b (mod m),
and if x0 x1 (mod m), then ax1 ax b (mod m),
So the entire congruence class will be a solution.
Congruences
Theorem 4.10 Let a, b and m be integers
such that m > 0 and (a,m) = d. If d | b,
then ax b (mod m) has no solutions.
If d |b, then ax b (mod m) has exactly d
incongruent solutions modulo m.
The solutions have the form:
x = x0 + (m/d)t , where t = 0, 1, 2, …, d-1
and x0 is a particular solution of ax – my = b.
Congruences
Let’s solve the following linear congruences.
a) 3x 6 (mod 9)
b) 15x 9 (mod 25)
c) 987x 610 (mod 1597)
Congruences
Corollary 4.10.1
If a and m are relatively prime integers, with
m > 0, and b an integer, then the linear
congruence ax b (mod m), has a unique solution
modulo m.
Example: Solve the linear congruence
17x 14 (mod 21)
Congruences
Def: Given an integer a with (a,m) = 1, a
solution of the linear congruence
ax 1 (mod m)
Is called an inverse of a modulo m.
Examples:
1) Since 7*3 = 21 1 (mod 20), we say that all
integers congruent to 3 modulo 20 are inverses of 7
modulo 20, 3 and 7 are inverses modulo 20.
2) How do we find inverses? Find the inverse of 6 mod 5
Congruences
Inverses allow us to solve linear congruence
Equations of the form: ax b (mod m)
Note: If a’ is an inverse of a modulo m, then we
know that
1) a*a’ 1 (mod m), so we have that
2) a’*ax a’b (mod m) which implies that
x a’b (mod m)
(So, you may use this technique to solve a linear
congruence, provided that you can easily access
the inverse of a.)
Congruences
Solve the following linear congruence:
19x 30 (mod 40)
Congruences
Theorem 4.11
Let p be a prime. The positive integer a is
its own inverse modulo p iff
a 1 (mod p) or a -1 (mod p).
Proof is BEAUTIFUL! Check it out!
Examples: Let p = 19. Then what are the
integers which are their own inverses?
Congruences
Next we want to see if we can solve
systems of linear congruences.
The origins of this are in “Mater Sun’s
Mathematical Manual” including many
chinese puzzles of the form:
“Find a number which leaves 1 when
divided by 2, 2 when divided by 3 and 3
when divided by 5.”
Congruences
Chinese Remainder Theorem:
Let m1, m2, …, mr be pairwise relatively prime
positive integers. Then the system of
Congurences
x a1 (mod m1)
x a2 (mod m2)
…
x ar (mod mr)
Has a unique solution modulo M = m1m2 ···mr.
Congruences
The solution of the above system has the
form:
x = a1 M1 y1 + a2 M2 y2 + … + ar Mr yr
Where Mi = M/mi and yi is the inverse of
Mi modulo mi.
Congruences
Let’s now solve the puzzle we started with
!
“Find a number which leaves 1 when
divided by 2, 2 when divided by 3 and 3
when divided by 5.”
Congruences
Alternatively, we can also solve such
systems using a sort of substitution
method. Let’s look at our puzzle:
Congruences
References:
Elementary Number Theory
And its applications
Fifth Edition
Kenneth H. Rosen