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  • What is linear congruence with example?

    A congruence of the form ax?b(mod m) where x is an unknown integer is called a linear congruence in one variable. It is important to know that if x0 is a solution for a linear congruence, then all integers xi such that xi?x0(mod m) are solutions of the linear congruence.

  • Different Methods to Solve Linear Congruences

    Example: Solve the linear congruence ax = b (mod m)Solution: ax = b (mod m) _____ (1)Example: Solve the linear congruence 3x = 12 (mod 6)Solution:Example: Solve the Linear Congruence 11x = 1 mod 23.Solution: Find the Greatest Common Divisor of the algorithm.

Simultaneous Linear, and Non-linear

Congruences

CIS002-2 Computational Alegrba and Number

Theory

David Goodwin

david.goodwin@perisic.com

09:00, Friday 24

thNovember 2011

09:00, Tuesday 28

thNovember 2011

09:00, Friday 02

ndDecember 2011 bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Outline

1Linear Congruences

2Simultaneous Linear Congruences

3Simultaneous Non-linear Congruences

4Chinese Remainder Theorem - An Extension

bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Outline

1Linear Congruences

2Simultaneous Linear Congruences

3Simultaneous Non-linear Congruences

4Chinese Remainder Theorem - An Extension

bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Theorem (5.6)

If d=gcd(a;n), then the linear congruence

axbmod (n) has a solution if and only if djb. If d does divide b, and if x0is any solution, then the general solution is given by x=x0+ntd where t2Z; in particular, the solutions form exactly d congruence classesmod(n), with representatives x=x0;x0+nd ;x0+2nd ;:::;x0+(d1)nd bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Lemma (5.7)

aLet mja;b;n, and let a0=a=m, b0=b=m and n0=n=m; then axbmod (n)if and only if a0xb0mod (n0) bLet a and n be coprime, let mja;b, and let a0=a=m and b

0=b=m; then

axbmod (n)if and only if a0xb0mod (n) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Algorithm for solution

1Calculated=gcd(a;n) and usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0sogcd(a00;b000)>1 and return to step

4 withb000instead ofb00. Or useca00xcb00mod (n0) in step

4, where the least absolute residea000ofca000satises

ja000jLinear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7), gcd(5;3) = 1,

5x3 mod (7),

56=1,

10 = 3 + (17)

gives 5x10 mod (7), gcd(5;10) = 5, x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,

5x3 mod (7),

56=1,

10 = 3 + (17)

gives 5x10 mod (7), gcd(5;10) = 5, x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7), 56=1,

10 = 3 + (17)

gives 5x10 mod (7), gcd(5;10) = 5, x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,

10 = 3 + (17)

gives 5x10 mod (7), gcd(5;10) = 5, x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,10 = 3 + (17) gives 5x10 mod (7), gcd(5;10) = 5, x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,10 = 3 + (17) gives 5x10 mod (7),gcd(5;10) = 5, x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,10 = 3 + (17) gives 5x10 mod (7),gcd(5;10) = 5,x2 mod (7), x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,10 = 3 + (17) gives 5x10 mod (7),gcd(5;10) = 5,x2 mod (7),x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,10 = 3 + (17) gives 5x10 mod (7),gcd(5;10) = 5,x2 mod (7),x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:10x6 mod (14)Example

gcd(10;14) = 2,5x3 mod (7),gcd(5;3) = 1,5x3 mod (7),56=1,10 = 3 + (17) gives 5x10 mod (7),gcd(5;10) = 5,x2 mod (7),x

0= 2,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 2 + 7t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:4x13 mod (47)Example

gcd(4;47) = 1,4x13 mod (47), 46=1,

412 = 481 mod (47)

x1213 mod (47) x3413 mod (47), x352 mod (47), x35 mod (47), x15 mod (47), x

0= 15,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 15 + 47t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:4x13 mod (47)Example

gcd(4;47) = 1,4x13 mod (47),46=1,

412 = 481 mod (47)

x1213 mod (47) x3413 mod (47), x352 mod (47), x35 mod (47), x15 mod (47), x

0= 15,1Calculated=gcd(a;n) and

usef0=fd

2Usea0xb0mod (n0)3Findm=gcd(a0;b0) and

usef00=fd

4Usea00xb00mod (n0)5Ifa00=1 thenx0=b006Elseuseb000=b00+kn0and

return to step 4. Or use ca

00xcb00mod (n0) and

return to step 4.So the general solution has the form x= 15 + 47t(t2Z) bg=white

Linear CongruencesSimul taneousLinear Cong ruencesSimul taneousNon-linear Congr uencesChinese Remainder Theorem - An Extension

Example:4x13 mod (47)Example

gcd(4;47) = 1,4x13 mod (47), 46=1,

412 = 481 mod (47)

x1213 mod (47) x3413 mod (47), x352 mod (47), x35 mod (47), x15 mod (47),quotesdbs_dbs12.pdfusesText_18
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