Mathematics 55– Spring 2005
Lecture 10 (Wednesday 2/9/2005)
Euclidean Algorithm and Congruence
Announcement: Midterm exam in class next Wednesday. (I supply paper.) Review
next Tuesday in discussion section.
Today: Congruences, congruence equations, Chinese Remainder Theorem. State-
ment of Fermat’s little Theorem.
Problem: Given a, b, how to solve the congruence equation ax ≡bmod m?
This should remind us of 9th grade algebra: to solve ax =b, we divide through
by ato get x=b/a =ba−1. What’s the analogue here? If we had a number ¯asuch
that ¯aa ≡1 mod mthen x= ¯ab would be a solution, for ax = (a¯a)b, and a¯a≡1
mod mimplies (by one of our theorems) that (a¯a)b≡bmod m. Such a number ¯a
is called a multiplicative inverse of amodulo m.
Of course, there are other solutions: if xis a solution then so is x+km for any
integer k; that is, if x≡ymod mthen yis also a solution. Reversing the reasoning,
we see that if ax ≡bmod mthen x≡¯ab mod m. Thus any two solutions must be
congruent modulo m; the solution is unique modulo m.
But when does ¯aexist? We know that there exist integers s, t such that gcd(a, m) =
sa+tm. If a, m are relatively prime, that is, if gcd(a, m) = 1, then we get 1 = sa+tm.
This says that sa ≡1 mod m, so ¯a=sis the desired multiplicative inverse of a.
On the other hand, if gcd(a, m) = cis 6= 1 then cdivides both a, m, so c|ax and
c|m, so cmust divide bin order for the equation to be solvable.
Upshot: If a, m are relatively prime then
(i) ahas a multiplicative inverse modulo m,
(ii) for any b, there is an integer solution xof the congruence equation ax ≡bmod m,
and
(iii) Any two solutions x, y are congruent modulo m.
Chinese Remainder Problem: Given finitely many (positive) integers m1, m2,· · · , mN,
and given integers b1,· · · , bN, find an integer xsatisfying x≡bjmod mjfor all
j∈ {1,2,· · · , N }simultaneously.
Example: Find xcongruent to 1 modulo 2, congruent to 2 modulo 3, and congruent
to 4 modulo 7. We check that x= 11 works; I didn’t explain how xcould be found.
Example: Find xcongruent to 1 modulo 8 and to 2 modulo 12. This is impossible;
the first congruence requires xto be odd, while the second requires it to be even.
Chinese Remainder Theorem: Let N≥2, and let m1,· · · , mN∈Nbe given. If
the numbers mjare pairwise relatively prime then for any numbers bj,
(i) a solution xexists,
(ii) any two solutions are congruent modulo Mwhere Mis the product M=
m1m2· · · mN.
(iii) if xis a solution, and if y≡xmod M, then yis also a solution.
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