3.1: Introduction to Congruences
As we mentioned in the introduction, the theory of congruences was developed by Gauss at the beginning of the nineteenth century.
Let \(m\) be a positive integer. We say that \(a\) is congruent to \(b\) modulo \(m\) if \(m \mid (a-b)\) where \(a\) and \(b\) are integers, i.e. if \(a=b+km\) where \(k\in \mathbb{Z}\).
If \(a\) is congruent to \(b\) modulo \(m\), we write \(a\equiv b(mod\ m)\).
\(19\equiv 5 (mod \ 7)\). Similarly \(2k+1 \equiv 1 (mod\ 2)\) which means every odd number is congruent to 1 modulo 2.
There are many common properties between equations and congruences. Some properties are listed in the following theorem.
Theorem: Properties of Congruences
Let \(a, b, c\) and \(d\) denote integers. Let \(m\) be a positive integers. Then:
- If \(a \equiv b(mod \ m)\) , then \(b\equiv a (mod \ m)\).
- If \(a\equiv b(mod \ m)\) and \(b\equiv c(mod \ m)\), then \(a\equiv c (mod \ m)\).
- If \(a\equiv b(mod\ m)\) , then \(a+c \equiv b+c (mod \ m)\).
- If \(a\equiv b(mod\ m)\) , then \(a-c \equiv b-c (mod \ m)\).
- If \(a\equiv b(mod\ m)\) , then \(ac \equiv bc (mod \ m)\).
- If \(a\equiv b(mod\ m)\) , then \(ac \equiv bc (mod \ mc)\), for \(c>0\).
- If \(a\equiv b(mod\ m)\) and \(c \equiv d (mod \ m)\) then \(a+c \equiv (b+d) (mod \ m)\).
- If \(a\equiv b(mod\ m)\) and \(c \equiv d (mod \ m)\) then \(a-c \equiv (b-d) (mod \ m)\).
- If \(a\equiv b(mod\ m)\) and \(c \equiv d (mod \ m)\) then \(ac \equiv bd (mod \ m)\).
- If \(a \equiv b(mod \ m)\) , then \(m\mid (a-b)\). Thus there exists integer \(k\) such that \(a-b=mk\), this implies \(b-a=m(-k)\) and thus \(m\mid (b-a)\). Consequently \(b\equiv a (mod \ m)\).
- Since \(a\equiv b(mod \ m)\) , then \(m\mid (a-b)\). Also, \(b\equiv c(mod \ m)\), then \(m\mid (b-c)\). As a result, there exit two integers \(k\) and \(l\) such that \(a=b+mk\) and \(b=c+ml\), which imply that \(a=c+m(k+l)\) giving that \(a=c (mod \ m)\).
- Since \(a\equiv b (mod \ m)\) , then \(m \mid (a-b)\). So if we add and subtract \(c\) we get \[m\mid ((a+c)-(b+c))\] and as a result \[a+c\equiv b+c (mod \ m).\]
- Since \(a\equiv b (mod \ m)\) , then \(m \mid (a-b)\) so we can subtract and add \(c\) and we get \[m\mid ((a-c)-(b-c))\] and as a result \[a-c\equiv b-c (mod \ m).\]
- If \(a \equiv b(mod \ m)\) , then \(m\mid (a-b)\). Thus there exists integer \(k\) such that \(a-b=mk\) and as a result \(ac-bc=m(kc)\). Thus \[m\mid (ac-bc)\] and hence \[ac\equiv bc (mod \ m).\]
- If \(a \equiv b(mod \ m)\) , then \(m\mid (a-b)\). Thus there exists integer \(k\) such that \(a-b=mk\) and as a result \[ac-bc=mc(k).\] Thus \[mc\mid (ac-bc)\] and hence \[ac\equiv bc (mod \ mc).\]
- Since \(a\equiv b(mod \ m)\) , then \(m\mid (a-b)\). Also, \(c\equiv d(mod \ m)\), then \(m\mid (c-d)\). As a result, there exits two integers \(k\) and \(l\) such that \(a-b=mk\) and \(c-d=ml\). Note that \[(a-b)+(c-d)=(a+c)-(b+d)=m(k+l).\] As a result, \[m\mid ((a+c)-(b+d)),\] hence \[a+c\equiv b+d(mod \ m).\]
- If \(a=b+mk\) and \(c=d+ml\) where \(k\) and \(l\) are integers, then \[(a-b)-(c-d)=(a-c)-(b-d)=m(k-l).\] As a result, \[m\mid ((a-c)-(b-d)),\] hence \[a-c\equiv b-d(mod \ m).\]
- There exit two integers \(k\) and \(l\) such that \(a-b=mk\) and \(c-d=ml\) and thus \(ca-cb=m(ck)\) and \(bc-bd=m(bl)\). Note that \[(ca-cb)+(bc-bd)=ac-bd=m(kc-lb).\] As a result, \[m\mid (ac-bd),\]hence \[ac\equiv bd(mod \ m).\]
- Because \(14\equiv 8(mod\ 6)\) then \(8 \equiv 14 (mod\ 6)\).
- Because \(22\equiv 10(mod \ 6)\) and \(10 \equiv 4(mod \ 6)\). Notice that \(22\equiv 4(mod \ 6)\).
- Because \(50\equiv 20 (mod\ 15)\) , then \(50+5=55\equiv 20+5=25(mod\ 15)\).
- Because \(50\equiv 20 (mod\ 15)\) , then \(50-5=45\equiv 20-5=15(mod\ 15)\).
- Because \(19\equiv 16(mod3)\) , then \(2(19)=38\equiv 2(16)=32(mod \ 3).\)
- Because \(19\equiv 16(mod3)\) , then \(2(19)=38\equiv 2(16)=32(mod \ 2(3)=6).\)
- Because \(19\equiv 3 (mod \ 8)\) and \(17\equiv 9(mod \ 8)\), then \(19+17=36\equiv 3+9=12(mod \ 8)\).
- Because \(19\equiv 3 (mod \ 8)\) and \(17\equiv 9(mod \ 8)\), then \(19-17=2\equiv 3-9=-6(mod \ 8)\).
- Because \(19\equiv 3 (mod \ 8)\) and \(17\equiv 9(mod \ 8)\), then \(19(17)=323\equiv 3(9)=27(mod \ 8)\).
We now present a theorem that will show one difference between equations and congruences. In equations, if we divide both sides of the equation by a non-zero number, equality holds. While in congruences, it is not necessarily true. In other words, dividing both sides of the congruence by the same integer doesn’t preserve the congruence.
- If \(a,b, c\) and \(m\) are integers such that \(m>0\), \(d=(m,c)\) and \(ac\equiv bc(mod \ m)\), then \(a\equiv b (mod \ m/d)\).
- If \((m,c)=1\) then \(a=b(mod \ m)\) if \(ac\equiv bc(mod \ m)\).
Part 2 follows immediately from Part 1. For Part 1, if \(ac\equiv bc(mod \ m)\), then \[m\mid (ac-bc)=c(a-b).\] Hence there exists \(k\) such that \(c(a-b)=mk\). Dividing both sides by \(d\), we get \((c/d)(a-b)=k(m/d)\). Since \((m/d,c/d)=1\), it follows that \(m/d \mid (a-b)\). Hence \(a\equiv b (mod \ m/d)\).
\(38 \equiv 10 (mod\ 7)\). Since \((2,7)=1\) then \(19\equiv 5 (mod \ 7).\)
The following theorem combines several congruences of two numbers with different moduli.
If \[a\equiv b(mod \ m_1), a\equiv b(mod \ m_2),...,a\equiv b(mod \ m_t)\] where \(a,b,m_1,m_2,...,m_t\) are integers and \(m_1,m_2,...,m_t\) are positive, then \[a\equiv b(mod \ \langle m_1,m_2,...m_t\rangle)\]
Since \(a\equiv b (mod \ m_i)\) for all \(1\leq i\leq t\). Thus \(m_i \mid (a-b)\). As a result, \[\langle m_1,m_2,...,m_t\rangle \mid (a-b)\] (prove this as an exercise). Thus \[a\equiv b(mod \ \langle m_1,m_2,...m_t\rangle).\]
Exercises
- Determine whether 3 and 99 are congruent modulo 7 or not.
- Show that if \(x\) is an odd integer, then \(x^2\equiv 1(mod \ 8)\)
- Show that if \(a,b, m\) and \(n\) are integers such that \(m\) and \(n\) are positive, \(n \mid m\) and \(a\equiv b (mod \ m)\), then \(a\equiv b (mod \ n).\)
- Show that if \(a_i\equiv b_i(mod \ m)\) for \(i=1,2,...,n\), where \(m\) is a positive integer and \(a_i,b_i\) are integers for \(j=1,2,...,n\), then \(\sum_{i=1}^na_i\equiv \sum_{i=1}^nb_i(mod \ m)\)
- For which \(n\) does the expression \(1+2+...+(n-1)\equiv 0(mod \ n)\) holds.
Contributors and Attributions
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Dr. Wissam Raji, Ph.D., of the American University in Beirut. His work was selected by the Saylor Foundation’s Open Textbook Challenge for public release under a Creative Commons Attribution ( CC BY ) license.